Apply linear & quadratic equations

1. Revision of solving equations

Exercises

2. Revision of simultaneous equations

Points of intersection

Draw the graphs of the following equations on the same Cartesian plane. Use the graphs to find the points of intersection of the two lines.

y=x2+4x+3y=2x+6
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The points of intersection are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to plot the graphs of the two equations on the same Cartesian plane.


STEP: Plot the graph of y=x2+4x+3 on a Cartesian plane
[−2 points ⇒ 4 / 6 points left]

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: y=x2+4x+3.

To plot this graph, we need to find the coordinates of the points where the graph will cross the x- and y-axes. These points are the x- and y-intercepts, respectively. We will then use these points to plot the graph.

The y-intercept is the quantity in the equation which is not multiplied by x. For this equation, the y-intercept is 3. The coordinates of this point are (0;3).

The x-intercepts are the roots of the equation x2+4x+3=0. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 4. This is the coefficient of the middle term. The product of the two numbers must be 3. This is the product of the coefficient of the first term 1 and the value of the last term 3. The two numbers are 1 and 3. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x2+4x+30=x2+x+3x+3

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

0=x(x+1)+3(x+1)0=(x+3)(x+1)

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are x=1 and x=3. The coordinates of these roots are (1;0) and (3;0).

Using these x- and y-intercepts we will draw the graph:


STEP: Plot the graph of y=2x+6 on the same Cartesian plane
[−2 points ⇒ 2 / 6 points left]

As we did for the quadratic equation, we will first find the y-intercept. After that we will calculate root for the linear equation. We remember that the y-intercept is the quantity in the equation which is not multiplied by x.

The y-intercept is 6. Its coordinates are: (0;6).

We will work out the root of the equation 0=2x+6.

0=2x+6=02x=6x=3

The root is: x=3. It's coordinates are: (3;0).

We will now plot the equation y=2x+6 on the same Cartesian plane we have used for the graph of the quadratic equation.


STEP: Read off the coordinates of the points of intersection
[−2 points ⇒ 0 / 6 points left]

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: (1;8) and (3;0).

We have used a graphical method to determine the solutions of the simultaneous equations: y=x2+4x+3 and y=2x+6. The solutions are the points where their graphs intersect.

NOTE: It is possible to solve the simultaneous equations using either the substitution or elimination methods.

The points of intersection are (1;8) and (3;0).


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Simultaneous equations: solving by substitution

Solve simultaneously for x and y. You should get two coordinate pairs, (x;y), for your answers.

y+4=x18x2+y+16=0
INSTRUCTION: When you type your answers they should look something like this: ( 2; -4 ) and ( -1; 3). It does not matter which coordinate pair you type first.
Answer: The solutions are and .
coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.


STEP: Rearrange the linear equation to isolate a variable
[−1 point ⇒ 5 / 6 points left]

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate y.

y+4=x18y=x22

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 6 points left]

Now we need to substitute the entire right side of the equation, (x22), into the quadratic equation in place of y. Be sure to use brackets! Then simplify the equation to get standard form.

x2+y+16=0x2+(x22)+16=0x2x6=0

STEP: Solve for the values of x
[−1 point ⇒ 2 / 6 points left]

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two x-values.

x2x6=0(x3)(x+2)=0x=3 and x=2

STEP: Find the value of y for each of the x-values
[−2 points ⇒ 0 / 6 points left]

Cool! However, there is still work to do. We must now find the values of y for these equations. We must substitute in the values x=3 and x=2. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=3:y=(3)22y=25If x=2:y=(2)22y=20

Finally we have the answers: if x=3 then y=25 and if x=2 then y=20.

NOTE: The graph below is not required for the solution. It is here to show the connection between the answers and the equations.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: (3;25) and (2;20).


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Simultaneous equations

Solve for x and y from the given equations:

y=x2+2xy=x+2
INSTRUCTION: You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided. It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x+2 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x2+2xlinear equationy=x+2

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

x+2=x2+2x0=x2+2xx20=x2+2xx20=x2+x2

STEP: Solve the equation x2+x2=0
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 1. This is the coefficient of the middle term. The product of the two numbers must be 2. This is the product of the coefficient of the first term 1 and the value of the last term 2. The two numbers are 2 and 1. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x2+x20=x2+2xx2

We now group the first two terms together and group the last terms together and factorise.

0=x2+2xx20=x(x+2)(x+2)

The last thing now is to factor the highest common factor.

0=(x1)(x+2)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 1 and 2.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

For x=1:

y=x+2=3

For x=2:

y=x+2=0

The solutions are x=1 together with y=3 , and x=2 with y=0. In coordinates pairs, the solutions are (1;3) and (2;0).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (1;3) and (2;0). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (1;3) and (2;0).


Submit your answer as: and

Simultaneous equations

Determine the solution of the following equations:

y=2x2+8x+7y=x+4
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x+4 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=2x2+8x+7linear equationy=x+4

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
x+4=2x2+8x+70=2x2+8x+7x40=2x2+8xx+740=2x2+7x+3

STEP: Solve the equation 0=2x2+7x+3
[−2 points ⇒ 2 / 6 points left]

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 7. This is the coefficient of the middle term. The product of the two numbers must be 6. This is the product of the coefficient of the first term 2 and the value of the last term 3. The two numbers are 1 and 6. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=2x2+7x+30=2x2+x+6x+3

We now group the first two terms together and group the last terms together and factorise.

0=2x2+x+6x+30=x(2x+1)+3(2x+1)

The last thing now is to factor the highest common factor.

0=(x+3)(2x+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 12 and 3.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.

For x=12:

y=x+4=72

For x=3:

y=x+4=1

The solutions are x=12 together with y=72 , and x=3 with y=1. In coordinate pairs, the solutions are (12;72) and (3;1).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (12;72) and (3;1). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (12;72) and (3;1).


Submit your answer as: and

Simultaneous equations

Solve for x and y:

y=x2+2x1x=y1
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y1 in place of x in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x2+2x1linear equationx=y1

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of x from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
y=(y1)2+2(y1)10=(y1)2+2(y1)1y0=y22y+1+(2y2)1y0=y22y+1+2y21y0=y22y+2yy+1210=y2y+1210=y2y110=y2y2

STEP: Solve the equation 0=y2y2
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 1. This is the coefficient of the middle term. The product of the two numbers must be 2. This is the product of the coefficient of the first term 1 and the value of the last term 2. The two numbers are 1 and 2. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=y2y20=y2+y2y2

We now group the first two terms together and group the last terms together and factorise.

0=y2+y2y20=y(y+1)2(y+1)

The last thing now is to factor the highest common factor.

0=(y2)(y+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are 2 and -1.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

We will substitute each y-value into the linear equation. This will give us the corresponding values of x.

NOTE: We can do that by substituting y into any of the original equations. It is easier to use the linear one.

For y=2:

x=y1=1

For y=1:

x=y1=2

The solutions are x=1 together with y=2 , and x=2 with y=1. In coordinates pairs, the solutions are (1;2) and (2;1).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (1;2) and (2;1). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (1;2) and (2;1).


Submit your answer as: and

Simultaneous equations: intersection points

The graph below shows the equations:

y=x23andy=x22+x4

Determine the points of intersection for these two curves.

INSTRUCTION:
  • Write your answers as coordinate pairs like this: ( 3; -4) and ( 1; 8).
  • It does not matter which coordinate pair you type first.
  • Do not round your answers.
Answer: The points of intersection are and .
coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Solve the equations simultaneously to find their intersection points.


STEP: Combine the equations by substitution
[−1 point ⇒ 5 / 6 points left]

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation y is equal to the quantity x23; and at the same time, the quadratic equation also contains the value y. Substitute like this:

y is x23and:yis also x22+x4Therefore: x23=x22+x4

STEP: Arrange the equation in standard form
[−2 points ⇒ 3 / 6 points left]

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is 2. So multiply both sides of the equation by 2 in order to cancel the denominators. Then arrange the equation in standard form.

(2)(x23)=(x22+x4)(2)x6=x2+2x80=x2+x2


STEP: Solve the equation
[−1 point ⇒ 2 / 6 points left]

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two x-values.

0=x2+x20=(x1)(x+2)x=2 and x=1

STEP: Use the values of x to find the values of y
[−2 points ⇒ 0 / 6 points left]

Great... but it isn't over yet. We must now find the values of y for these equations. Substitute in the values x=2 and x=1 into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=2:y=(2)23y=4If x=1:y=(1)23y=52

Finally! The solutions are (2;4) and (1;52).

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are (2;4) and (1;52).


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Substitution with quadratic equations

Here are two equations, which we can solve simultaneously using substitution:

y=x2+2x1x=y1

Which of the following equations shows the correct substitution step?

A y=(y+1)2+2(y+1)1
B y=(y1)2+2(y1)1
C y=y2+2y1
D y=y12+2y11
Answer: The correct substitution step is choice: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to substitute y1 in place of x in the quadratic equation y=x2+2x1.


STEP: Substitute the linear equation into the quadratic equation
[−1 point ⇒ 0 / 1 points left]

We have two equations that we will solve by substitution. The equations are:

quadratic equationy=x2+2x1linear equationx=y1

We need to substitute y1 from the linear equation in place of x in the quadratic equation.

y=(y1)2+2(y1)1
NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. The solution for this question did not consider this case.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for y. The question does not ask us to solve the simultaneous equations.

The correct substitution step is y=(y1)2+2(y1)1 which is choice B.


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Equations with fractions

Determine the values of x and y which solve these two equations:

3xy=103x=4y20xyx,y0

Your answer should be exact (do not round off).

Answer:

The solution is x= and y= .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The first thing you should do is rearrange the second equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is xy.


STEP: Rearrange the second equation
[−1 point ⇒ 4 / 5 points left]

We need to solve the equations simultaneously. To do that, we should start by rearranging the second equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by xy, which is the LCD of the fractions. That will cancel all of the denominators.

3x=4y20xyxy(3x)=xy(4y20xy)3xxy=4yxy203y=4x20

STEP: Pick a method and solve for whichever variable comes more easily
[−3 points ⇒ 1 / 5 points left]

Now we are ready to solve the following equations:

3xy=103y=4x20

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations are not arranged nicely for elimination or substitution. We must change the equations somehow. In this case, we can make a small change that will allow us to use elimination: we can multiply the first equation by 3 so that the y-coefficients will be ready to cancel. Then we can eliminate them by subtracting the equations.

First modify things to set up the elimination:

3(3xy)=3(10)9x3y=30

Now do the elimination and complete the solution to find x.

9x3y=30second equationSubtract the(3y)=(4x20)9x3y+3y=30+20+4x9x=4x105x=10x=2

Great - we have the first value, x=2.


STEP: Solve for the other variable
[−1 point ⇒ 0 / 5 points left]

Now the last step: find the value of y. We can do this using either of the equations in the question: but the first eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the second equation).

3xy=103(2)y=10y=4y=4

The answer is the pair of numbers x=2 and y=4. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

NOTE: Remember, the fractions force x,y0, which is why there are open intervals at the x- and y-intercepts of the fraction equation.

The values which solve these two equations are x=2 and y=4.


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Simultaneous equations: special outcomes

Solve for x and y from the given equations:

y=3x2+3x+2y=x5
INSTRUCTION: You should enter your answer in the form of a coordinate pair, (x; y). An example of an acceptable answer is (2; 5). If there is no solution, type no solution.
Answer: The solution is .
one-of
type(string.nocase)
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to start by substituting x5 in place of y in the quadratic equation and simplify. You must expect a special kind of solution.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 2 / 4 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is y=3x2+3x+2 and the linear equation is y=x5. This means we expect two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
x5=3x2+3x+20=3x2+3x+2+x+50=3x2+3x+x+2+50=3x2+4x+7

STEP: Solve the equation 0=3x2+4x+7
[−2 points ⇒ 0 / 4 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

We need two numbers that add to give 4. This is the coefficient of the middle term. The product of the two numbers must be 21. This is the product of the coefficient of the first term 3 and the value of the last term 7. In this case, there are no such numbers. We will now use the quadratic formula to solve this equation.

The solution for a quadratic equation of the form ax2+bx+c=0:

x=b±(b24ac)2a

Before we can use the quadratic formula, we need to identify the coefficients a, b and c from the equation 3x2+4x+7=0 These are:

a=3,b=4 and c=7.

We will substitute these values into the formula and simplify.

x=(4)±((4)24(3)(7))2(3)=4±(16(84))6=4±(68)6

The number under the square root sign is negative. This means that our quadratic equation has no real solution. This means that y=3x2+3x+2 and y=x5 cannot be solved simultaneously.

We can confirm our result by plotting the graph of each equation on the same set of axes. We do not expect these graphs to intersect.

Aha! The two graphs do not have an intersection point. This means that indeed the two equations have no solution.

The final answer is no solution.


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Simultaneous equations

Adapted from DBE Nov 2016 Grade 12, P1, Q1.3
Maths formulas

Solve for x and y:

x=2yandx2+3xy=18
INSTRUCTIONS:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are
and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.


STEP: Substitute x=2y into x2+3xy=18 and solve for y
[−3 points ⇒ 1 / 4 points left]

We need to solve the following pair of equations for x and y:

x=2yandx2+3xy=18

First, we can substitute 2y in place of x in the second equation:

(2y)2+3(2y)(y)=18

We can simplify this equation to solve for y:

(2y)2+3(2y)(y)=184y26y2=182y2=18y2=9

y can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about y2. So we have to consider both the positive and the negative values as solutions to the equations.

y=3 or y=3

STEP: Use the y-values to find the x-values
[−1 point ⇒ 0 / 4 points left]

Each of the y-values that we have found will have a corresponding value of x, which can be found from the equation x=2y.

For y=3:

x=2y=2(3)=6

And for y=3:

x=2y=2(3)=6

So the x-values are x=6 or x=6.

Therefore the correct coordinate pairs are (6;3) and (6;3).


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Simultaneous equations with xy terms

Solve for x and y in the following simultaneous equations:

y=9xxy+1y=x+3
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x+3 in place of y in the hyperbolic equation and then solve for x.


STEP: Substitute the linear equation into the hyperbolic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

hyperbolic equationy=9xxy+1linear equationy=x+3

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

NOTE: We could also isolate either x or y from y=9xxy+1 and substitute its value into y=x+3. We could still get the same answer, but the calculations might become too long.
(x+3)=9xx(x+3)+10=9xx(x+3)+1+(x+3)0=9x(x2+3x)+1+(x+3)0=x2+6xx+1+30=x2+5x+1+30=x2+5x+4

STEP: Solve the equation 0=x2+5x+4
[−2 points ⇒ 2 / 6 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x2+5x+40=(x+1)(x+4)

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 1 and 4.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

By using the values of x, we will calculate the corresponding values of y. We will do so by substituting each x-value into the linear equation.

NOTE: We can still get the same values of y if we substitute the values x into the equation y=9xxy+1. If we do that, the calculations might become too long.

For x=1:

y=x+3=4

For x=4:

y=x+3=7

The solutions are x=1 together with y=4 , and x=4 with y=7. Using coordinate pairs, the solutions are (1;4) and (4;7).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (1;4) and (4;7). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (1;4) and (4;7).


Submit your answer as: and

Exercises

Points of intersection

Draw the graphs of the following equations on the same Cartesian plane. Use the graphs to find the points of intersection of the two lines.

y=x2+4x+3y=x1
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The points of intersection are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to plot the graphs of the two equations on the same Cartesian plane.


STEP: Plot the graph of y=x2+4x+3 on a Cartesian plane
[−2 points ⇒ 4 / 6 points left]

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: y=x2+4x+3.

To plot this graph, we need to find the coordinates of the points where the graph will cross the x- and y-axes. These points are the x- and y-intercepts, respectively. We will then use these points to plot the graph.

The y-intercept is the quantity in the equation which is not multiplied by x. For this equation, the y-intercept is 3. The coordinates of this point are (0;3).

The x-intercepts are the roots of the equation x2+4x+3=0. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 4. This is the coefficient of the middle term. The product of the two numbers must be 3. This is the product of the coefficient of the first term 1 and the value of the last term 3. The two numbers are 1 and 3. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x2+4x+30=x2+x+3x+3

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

0=x(x+1)+3(x+1)0=(x+3)(x+1)

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are x=1 and x=3. The coordinates of these roots are (1;0) and (3;0).

Using these x- and y-intercepts we will draw the graph:


STEP: Plot the graph of y=x1 on the same Cartesian plane
[−2 points ⇒ 2 / 6 points left]

As we did for the quadratic equation, we will first find the y-intercept. After that we will calculate root for the linear equation. We remember that the y-intercept is the quantity in the equation which is not multiplied by x.

The y-intercept is 1. Its coordinates are: (0;1).

We will work out the root of the equation 0=x1.

0=x1=0x=1x=1

The root is: x=1. It's coordinates are: (1;0).

We will now plot the equation y=x1 on the same Cartesian plane we have used for the graph of the quadratic equation.


STEP: Read off the coordinates of the points of intersection
[−2 points ⇒ 0 / 6 points left]

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: (1;0) and (4;3).

We have used a graphical method to determine the solutions of the simultaneous equations: y=x2+4x+3 and y=x1. The solutions are the points where their graphs intersect.

NOTE: It is possible to solve the simultaneous equations using either the substitution or elimination methods.

The points of intersection are (1;0) and (4;3).


Submit your answer as: and

Points of intersection

Sketch the graphs of the following equations on the same Cartesian plane. Use the graphs to find the points of intersection of the two lines.

y=x2+6x+5y=2x+2
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The points of intersection are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to plot the graphs of the two equations on the same Cartesian plane.


STEP: Plot the graph of y=x2+6x+5 on a Cartesian plane
[−2 points ⇒ 4 / 6 points left]

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: y=x2+6x+5.

To plot this graph, we need to find the coordinates of the points where the graph will cross the x- and y-axes. These points are the x- and y-intercepts, respectively. We will then use these points to plot the graph.

The y-intercept is the quantity in the equation which is not multiplied by x. For this equation, the y-intercept is 5. The coordinates of this point are (0;5).

The x-intercepts are the roots of the equation x2+6x+5=0. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 6. This is the coefficient of the middle term. The product of the two numbers must be 5. This is the product of the coefficient of the first term 1 and the value of the last term 5. The two numbers are 1 and 5. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x2+6x+50=x2+x+5x+5

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

0=x(x+1)+5(x+1)0=(x+5)(x+1)

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are x=1 and x=5. The coordinates of these roots are (1;0) and (5;0).

Using these x- and y-intercepts we will draw the graph:


STEP: Plot the graph of y=2x+2 on the same Cartesian plane
[−2 points ⇒ 2 / 6 points left]

As we did for the quadratic equation, we will first find the y-intercept. After that we will calculate root for the linear equation. We remember that the y-intercept is the quantity in the equation which is not multiplied by x.

The y-intercept is 2. Its coordinates are: (0;2).

We will work out the root of the equation 0=2x+2.

0=2x+2=02x=2x=1

The root is: x=1. It's coordinates are: (1;0).

We will now plot the equation y=2x+2 on the same Cartesian plane we have used for the graph of the quadratic equation.


STEP: Read off the coordinates of the points of intersection
[−2 points ⇒ 0 / 6 points left]

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: (1;0) and (3;4).

We have used a graphical method to determine the solutions of the simultaneous equations: y=x2+6x+5 and y=2x+2. The solutions are the points where their graphs intersect.

NOTE: It is possible to solve the simultaneous equations using either the substitution or elimination methods.

The points of intersection are (1;0) and (3;4).


Submit your answer as: and

Points of intersection

Draw the graphs of the following equations on the same Cartesian plane. Use the graphs to find the points of intersection of the two lines.

y=x2+4x+3y=x+7
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The points of intersection are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to plot the graphs of the two equations on the same Cartesian plane.


STEP: Plot the graph of y=x2+4x+3 on a Cartesian plane
[−2 points ⇒ 4 / 6 points left]

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: y=x2+4x+3.

To plot this graph, we need to find the coordinates of the points where the graph will cross the x- and y-axes. These points are the x- and y-intercepts, respectively. We will then use these points to plot the graph.

The y-intercept is the quantity in the equation which is not multiplied by x. For this equation, the y-intercept is 3. The coordinates of this point are (0;3).

The x-intercepts are the roots of the equation x2+4x+3=0. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 4. This is the coefficient of the middle term. The product of the two numbers must be 3. This is the product of the coefficient of the first term 1 and the value of the last term 3. The two numbers are 1 and 3. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x2+4x+30=x2+x+3x+3

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

0=x(x+1)+3(x+1)0=(x+3)(x+1)

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are x=1 and x=3. The coordinates of these roots are (1;0) and (3;0).

Using these x- and y-intercepts we will draw the graph:


STEP: Plot the graph of y=x+7 on the same Cartesian plane
[−2 points ⇒ 2 / 6 points left]

As we did for the quadratic equation, we will first find the y-intercept. After that we will calculate root for the linear equation. We remember that the y-intercept is the quantity in the equation which is not multiplied by x.

The y-intercept is 7. Its coordinates are: (0;7).

We will work out the root of the equation 0=x+7.

0=x+7=0x=7

The root is: x=7. It's coordinates are: (7;0).

We will now plot the equation y=x+7 on the same Cartesian plane we have used for the graph of the quadratic equation.


STEP: Read off the coordinates of the points of intersection
[−2 points ⇒ 0 / 6 points left]

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: (1;8) and (4;3).

We have used a graphical method to determine the solutions of the simultaneous equations: y=x2+4x+3 and y=x+7. The solutions are the points where their graphs intersect.

NOTE: It is possible to solve the simultaneous equations using either the substitution or elimination methods.

The points of intersection are (1;8) and (4;3).


Submit your answer as: and

Simultaneous equations: solving by substitution

Solve this pair of simultaneous equations. You should get two coordinate pairs, (x;y), for your answers.

x5=3y2x+5y=y2+6
INSTRUCTION: When you type your answers they should look something like this: ( 2; -4 ) and ( -1; 3). It does not matter which coordinate pair you type first.
Answer: The solutions are and .
coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.


STEP: Rearrange the linear equation to isolate a variable
[−1 point ⇒ 5 / 6 points left]

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate x.

x5=3y2x=3y+3

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 6 points left]

Now we need to substitute the entire right side of the equation, (3y+3), into the quadratic equation in place of x. Be sure to use brackets! Then simplify the equation to get standard form.

x+5y=y2+6(3y+3)+5y=y2+62y+3=y2+6y2+2y3=0to keep the quadratic term positivemove the terms onto the left side

STEP: Solve for the values of y
[−1 point ⇒ 2 / 6 points left]

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two y-values.

y2+2y3=0(y1)(y+3)=0y=1 and y=3

STEP: Find the value of x for each of the y-values
[−2 points ⇒ 0 / 6 points left]

Cool! However, there is still work to do. We must now find the values of x for these equations. We must substitute in the values y=1 and y=3. We can use either equation to do this, so we will choose the easier option: the linear equation.

If y=1:x=3(1)+3x=0If y=3:x=3(3)+3x=12

Finally we have the answers: if x=0 then y=1 and if x=12 then y=3.

NOTE: The graph below is not required for the solution. It is here to show the connection between the answers and the equations.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: (0;1) and (12;3).


Submit your answer as: and

Simultaneous equations: solving by substitution

Solve simultaneously for x and y. You should get two coordinate pairs, (x;y), for your answers.

3x+y+16=00=x2y16
INSTRUCTION: When you type your answers they should look something like this: ( 2; -4 ) and ( -1; 3). It does not matter which coordinate pair you type first.
Answer: The solutions are and .
coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.


STEP: Rearrange the linear equation to isolate a variable
[−1 point ⇒ 5 / 6 points left]

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate y.

3x+y+16=0y=3x16

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 6 points left]

Now we need to substitute the entire right side of the equation, (3x16), into the quadratic equation in place of y. Be sure to use brackets! Then simplify the equation to get standard form.

0=x2y160=x2(3x16)160=x23x

STEP: Solve for the values of x
[−1 point ⇒ 2 / 6 points left]

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two x-values.

0=x23x0=x(x3)x=3 and x=0

STEP: Find the value of y for each of the x-values
[−2 points ⇒ 0 / 6 points left]

Cool! However, there is still work to do. We must now find the values of y for these equations. We must substitute in the values x=3 and x=0. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=3:y=3(3)16y=7If x=0:y=3(0)16y=16

Finally we have the answers: if x=3 then y=7 and if x=0 then y=16.

NOTE: The graph below is not required for the solution. It is here to show the connection between the answers and the equations.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: (3;7) and (0;16).


Submit your answer as: and

Simultaneous equations: solving by substitution

Find the complete solution for these simultaneous equations. You should get two coordinate pairs, (x;y), for your answers.

0=xy+20y28=x+6y
INSTRUCTION: When you type your answers they should look something like this: ( 2; -4 ) and ( -1; 3). It does not matter which coordinate pair you type first.
Answer: The solutions are and .
coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.


STEP: Rearrange the linear equation to isolate a variable
[−1 point ⇒ 5 / 6 points left]

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate x.

0=xy+20x=y+20

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 6 points left]

Now we need to substitute the entire right side of the equation, (y+20), into the quadratic equation in place of x. Be sure to use brackets! Then simplify the equation to get standard form.

y28=x+6yy28=(y+20)+6yy28=7y20y27y+12=0to keep the quadratic term positivemove the terms onto the left side

STEP: Solve for the values of y
[−1 point ⇒ 2 / 6 points left]

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two y-values.

y27y+12=0(y4)(y3)=0y=4 and y=3

STEP: Find the value of x for each of the y-values
[−2 points ⇒ 0 / 6 points left]

Cool! However, there is still work to do. We must now find the values of x for these equations. We must substitute in the values y=4 and y=3. We can use either equation to do this, so we will choose the easier option: the linear equation.

If y=4:x=(4)+20x=16If y=3:x=(3)+20x=17

Finally we have the answers: if x=16 then y=4 and if x=17 then y=3.

NOTE: The graph below is not required for the solution. It is here to show the connection between the answers and the equations.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: (16;4) and (17;3).


Submit your answer as: and

Simultaneous equations

Solve for x and y in the following simultaneous equations:

y=x24xy=x4
INSTRUCTION: You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided. It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x4 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x24xlinear equationy=x4

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

x4=x24x0=x24xx+40=x24xx+40=x25x+4

STEP: Solve the equation x25x+4=0
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 5. This is the coefficient of the middle term. The product of the two numbers must be 4. This is the product of the coefficient of the first term 1 and the value of the last term 4. The two numbers are 1 and 4. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x25x+40=x2x4x+4

We now group the first two terms together and group the last terms together and factorise.

0=x2x4x+40=x(x1)4(x1)

The last thing now is to factor the highest common factor.

0=(x4)(x1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 4 and 1.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

For x=4:

y=x4=0

For x=1:

y=x4=3

The solutions are x=4 together with y=0 , and x=1 with y=3. In coordinates pairs, the solutions are (4;0) and (1;3).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (4;0) and (1;3). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (4;0) and (1;3).


Submit your answer as: and

Simultaneous equations

Solve the following equations simultaneously:

y=x2+5x+4y=x+1
INSTRUCTION: You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided. It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x+1 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x2+5x+4linear equationy=x+1

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

x+1=x2+5x+40=x2+5x+4x10=x2+5xx+410=x2+4x+3

STEP: Solve the equation x2+4x+3=0
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 4. This is the coefficient of the middle term. The product of the two numbers must be 3. This is the product of the coefficient of the first term 1 and the value of the last term 3. The two numbers are 1 and 3. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x2+4x+30=x2+x+3x+3

We now group the first two terms together and group the last terms together and factorise.

0=x2+x+3x+30=x(x+1)+3(x+1)

The last thing now is to factor the highest common factor.

0=(x+3)(x+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 1 and 3.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

For x=1:

y=x+1=0

For x=3:

y=x+1=2

The solutions are x=1 together with y=0 , and x=3 with y=2. In coordinates pairs, the solutions are (1;0) and (3;2).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (1;0) and (3;2). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (1;0) and (3;2).


Submit your answer as: and

Simultaneous equations

Solve for x and y from the given equations:

y=3x210x+6y=2x3
INSTRUCTION: You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided. It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting 2x3 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=3x210x+6linear equationy=2x3

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

2x3=3x210x+60=3x210x+62x+30=3x210x2x+6+30=3x212x+9

STEP: Solve the equation 3x212x+9=0
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 12. This is the coefficient of the middle term. The product of the two numbers must be 27. This is the product of the coefficient of the first term 3 and the value of the last term 9. The two numbers are 3 and 9. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=3x212x+90=3x23x9x+9

We now group the first two terms together and group the last terms together and factorise.

0=3x23x9x+90=3x(x1)9(x1)

The last thing now is to factor the highest common factor.

0=(3x9)(x1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 3 and 1.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

For x=3:

y=2x3=3

For x=1:

y=2x3=1

The solutions are x=3 together with y=3 , and x=1 with y=1. In coordinates pairs, the solutions are (3;3) and (1;1).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (3;3) and (1;1). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (3;3) and (1;1).


Submit your answer as: and

Simultaneous equations

Solve for x and y from the given equations:

y=2x22x1y=x+2
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x+2 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=2x22x1linear equationy=x+2

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
x+2=2x22x10=2x22x1+x20=2x22x+x120=2x2x3

STEP: Solve the equation 0=2x2x3
[−2 points ⇒ 2 / 6 points left]

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 1. This is the coefficient of the middle term. The product of the two numbers must be 6. This is the product of the coefficient of the first term 2 and the value of the last term 3. The two numbers are 2 and 3. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=2x2x30=2x2+2x3x3

We now group the first two terms together and group the last terms together and factorise.

0=2x2+2x3x30=2x(x+1)3(x+1)

The last thing now is to factor the highest common factor.

0=(2x3)(x+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 32 and 1.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.

For x=32:

y=x+2=12

For x=1:

y=x+2=3

The solutions are x=32 together with y=12 , and x=1 with y=3. In coordinate pairs, the solutions are (32;12) and (1;3).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (32;12) and (1;3). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (32;12) and (1;3).


Submit your answer as: and

Simultaneous equations

Solve for x and y in the following simultaneous equations:

y=2x24x2y=x+3
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x+3 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=2x24x2linear equationy=x+3

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
x+3=2x24x20=2x24x2+x30=2x24x+x230=2x23x5

STEP: Solve the equation 0=2x23x5
[−2 points ⇒ 2 / 6 points left]

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 3. This is the coefficient of the middle term. The product of the two numbers must be 10. This is the product of the coefficient of the first term 2 and the value of the last term 5. The two numbers are 2 and 5. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=2x23x50=2x2+2x5x5

We now group the first two terms together and group the last terms together and factorise.

0=2x2+2x5x50=2x(x+1)5(x+1)

The last thing now is to factor the highest common factor.

0=(2x5)(x+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 52 and 1.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.

For x=52:

y=x+3=12

For x=1:

y=x+3=4

The solutions are x=52 together with y=12 , and x=1 with y=4. In coordinate pairs, the solutions are (52;12) and (1;4).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (52;12) and (1;4). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (52;12) and (1;4).


Submit your answer as: and

Simultaneous equations

Solve for x and y in the following simultaneous equations:

y=2x28x+6y=x1
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x1 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=2x28x+6linear equationy=x1

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
x1=2x28x+60=2x28x+6x+10=2x28xx+6+10=2x29x+7

STEP: Solve the equation 0=2x29x+7
[−2 points ⇒ 2 / 6 points left]

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 9. This is the coefficient of the middle term. The product of the two numbers must be 14. This is the product of the coefficient of the first term 2 and the value of the last term 7. The two numbers are 2 and 7. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=2x29x+70=2x22x7x+7

We now group the first two terms together and group the last terms together and factorise.

0=2x22x7x+70=2x(x1)7(x1)

The last thing now is to factor the highest common factor.

0=(2x7)(x1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 72 and 1.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.

For x=72:

y=x1=52

For x=1:

y=x1=0

The solutions are x=72 together with y=52 , and x=1 with y=0. In coordinate pairs, the solutions are (72;52) and (1;0).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (72;52) and (1;0). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (72;52) and (1;0).


Submit your answer as: and

Simultaneous equations

Solve for x and y in the following simultaneous equations:

y=3x25x+2x=y2
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y2 in place of x in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=3x25x+2linear equationx=y2

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of x from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
y=3(y2)25(y2)+20=(3)(y2)25(y2)+2y0=3(y24y+4)(5y10)+2y0=3y212y+12(5y10)+2y0=3y212y+125y+10+2y0=3y212y5yy+12+10+20=3y217yy+12+10+20=3y218y+12+10+20=3y218y+22+20=3y218y+24

STEP: Solve the equation 0=3y218y+24
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 18. This is the coefficient of the middle term. The product of the two numbers must be 72. This is the product of the coefficient of the first term 3 and the value of the last term 24. The two numbers are 6 and 12. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=3y218y+240=3y26y12y+24

We now group the first two terms together and group the last terms together and factorise.

0=3y26y12y+240=3y(y+2)+12(y+2)

The last thing now is to factor the highest common factor.

0=(3y+12)(y+2)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are 4 and 2.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

We will substitute each y-value into the linear equation. This will give us the corresponding values of x.

NOTE: We can do that by substituting y into any of the original equations. It is easier to use the linear one.

For y=4:

x=y2=2

For y=2:

x=y2=0

The solutions are x=2 together with y=4 , and x=0 with y=2. In coordinates pairs, the solutions are (2;4) and (0;2).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (2;4) and (0;2). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (2;4) and (0;2).


Submit your answer as: and

Simultaneous equations

Solve for x and y in the following simultaneous equations:

y=2x23x+1x=y1
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y1 in place of x in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=2x23x+1linear equationx=y1

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of x from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
y=2(y1)23(y1)+10=(2)(y1)23(y1)+1y0=2(y22y+1)(3y3)+1y0=2y24y+2(3y3)+1y0=2y24y+23y+3+1y0=2y24y3yy+2+3+10=2y27yy+2+3+10=2y28y+2+3+10=2y28y+5+10=2y28y+6

STEP: Solve the equation 0=2y28y+6
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 8. This is the coefficient of the middle term. The product of the two numbers must be 12. This is the product of the coefficient of the first term 2 and the value of the last term 6. The two numbers are 2 and 6. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=2y28y+60=2y22y6y+6

We now group the first two terms together and group the last terms together and factorise.

0=2y22y6y+60=2y(y1)6(y1)

The last thing now is to factor the highest common factor.

0=(2y6)(y1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are 3 and 1.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

We will substitute each y-value into the linear equation. This will give us the corresponding values of x.

NOTE: We can do that by substituting y into any of the original equations. It is easier to use the linear one.

For y=3:

x=y1=2

For y=1:

x=y1=0

The solutions are x=2 together with y=3 , and x=0 with y=1. In coordinates pairs, the solutions are (2;3) and (0;1).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (2;3) and (0;1). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (2;3) and (0;1).


Submit your answer as: and

Simultaneous equations

Solve for x and y in the following simultaneous equations:

y=x2+x+2x=y3
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y3 in place of x in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x2+x+2linear equationx=y3

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of x from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
y=(y3)2+(y3)+20=(y3)2+(y3)+2y0=y26y+9+(y3)+2y0=y26y+9+y3+2y0=y26y+yy+93+20=y25yy+93+20=y26y+93+20=y26y+6+20=y26y+8

STEP: Solve the equation 0=y26y+8
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 6. This is the coefficient of the middle term. The product of the two numbers must be 8. This is the product of the coefficient of the first term 1 and the value of the last term 8. The two numbers are 2 and 4. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=y26y+80=y22y4y+8

We now group the first two terms together and group the last terms together and factorise.

0=y22y4y+80=y(y+2)+4(y+2)

The last thing now is to factor the highest common factor.

0=(y+4)(y+2)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are 4 and 2.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

We will substitute each y-value into the linear equation. This will give us the corresponding values of x.

NOTE: We can do that by substituting y into any of the original equations. It is easier to use the linear one.

For y=4:

x=y3=1

For y=2:

x=y3=1

The solutions are x=1 together with y=4 , and x=1 with y=2. In coordinates pairs, the solutions are (1;4) and (1;2).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (1;4) and (1;2). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (1;4) and (1;2).


Submit your answer as: and

Simultaneous equations: intersection points

The graph below shows the equations:

y=8x3+323andy=4x23+4x+163

Determine the points of intersection for these two curves.

INSTRUCTION:
  • Write your answers as coordinate pairs like this: ( 3; -4) and ( 1; 8).
  • It does not matter which coordinate pair you type first.
  • Do not round your answers.
Answer: The points of intersection are and .
coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Solve the equations simultaneously to find their intersection points.


STEP: Combine the equations by substitution
[−1 point ⇒ 5 / 6 points left]

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation y is equal to the quantity 8x3+323; and at the same time, the quadratic equation also contains the value y. Substitute like this:

y is 8x3+323and:yis also 4x23+4x+163Therefore: 8x3+323=4x23+4x+163

STEP: Arrange the equation in standard form
[−2 points ⇒ 3 / 6 points left]

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is 3. So multiply both sides of the equation by 3 in order to cancel the denominators. Then arrange the equation in standard form.

(3)(8x3+323)=(4x23+4x+163)(3)8x+32=4x2+12x+160=4x2+20x16

In this case, there is a common factor of −4 for all of the terms. Divide out this factor before factorising.

0=x25x+4

STEP: Solve the equation
[−1 point ⇒ 2 / 6 points left]

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two x-values.

0=x25x+40=(x4)(x1)x=1 and x=4

STEP: Use the values of x to find the values of y
[−2 points ⇒ 0 / 6 points left]

Great... but it isn't over yet. We must now find the values of y for these equations. Substitute in the values x=1 and x=4 into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=1:y=8(1)3+323y=8If x=4:y=8(4)3+323y=0

Finally! The solutions are (1;8) and (4;0).

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are (1;8) and (4;0).


Submit your answer as: and

Simultaneous equations: intersection points

The graph below shows the equations:

y=8x3andy=4x234x316

Determine the points of intersection for these two curves.

INSTRUCTION:
  • Write your answers as coordinate pairs like this: ( 3; -4) and ( 1; 8).
  • It does not matter which coordinate pair you type first.
  • Do not round your answers.
Answer: The points of intersection are and .
coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Solve the equations simultaneously to find their intersection points.


STEP: Combine the equations by substitution
[−1 point ⇒ 5 / 6 points left]

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation y is equal to the quantity 8x3; and at the same time, the quadratic equation also contains the value y. Substitute like this:

y is 8x3and:yis also 4x234x316Therefore: 8x3=4x234x316

STEP: Arrange the equation in standard form
[−2 points ⇒ 3 / 6 points left]

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is 3. So multiply both sides of the equation by 3 in order to cancel the denominators. Then arrange the equation in standard form.

(3)(8x3)=(4x234x316)(3)8x=4x24x480=4x2+4x48

In this case, there is a common factor of 4 for all of the terms. Divide out this factor before factorising.

0=x2+x12

STEP: Solve the equation
[−1 point ⇒ 2 / 6 points left]

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two x-values.

0=x2+x120=(x3)(x+4)x=4 and x=3

STEP: Use the values of x to find the values of y
[−2 points ⇒ 0 / 6 points left]

Great... but it isn't over yet. We must now find the values of y for these equations. Substitute in the values x=4 and x=3 into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=4:y=8(4)3y=323If x=3:y=8(3)3y=8

Finally! The solutions are (4;323) and (3;8).

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are (4;323) and (3;8).


Submit your answer as: and

Simultaneous equations: intersection points

The graph below shows the equations:

y=4x343andy=4x2320x38

Find the intersection points of the curves.

INSTRUCTION:
  • Write your answers as coordinate pairs like this: ( 3; -4) and ( 1; 8).
  • It does not matter which coordinate pair you type first.
  • Do not round your answers.
Answer: The points of intersection are and .
coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Solve the equations simultaneously to find their intersection points.


STEP: Combine the equations by substitution
[−1 point ⇒ 5 / 6 points left]

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation y is equal to the quantity 4x343; and at the same time, the quadratic equation also contains the value y. Substitute like this:

y is 4x343and:yis also 4x2320x38Therefore: 4x343=4x2320x38

STEP: Arrange the equation in standard form
[−2 points ⇒ 3 / 6 points left]

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is 3. So multiply both sides of the equation by 3 in order to cancel the denominators. Then arrange the equation in standard form.

(3)(4x343)=(4x2320x38)(3)4x4=4x220x240=4x224x20

In this case, there is a common factor of −4 for all of the terms. Divide out this factor before factorising.

0=x2+6x+5

STEP: Solve the equation
[−1 point ⇒ 2 / 6 points left]

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two x-values.

0=x2+6x+50=(x+1)(x+5)x=5 and x=1

STEP: Use the values of x to find the values of y
[−2 points ⇒ 0 / 6 points left]

Great... but it isn't over yet. We must now find the values of y for these equations. Substitute in the values x=5 and x=1 into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=5:y=4(5)343y=8If x=1:y=4(1)343y=83

Finally! The solutions are (5;8) and (1;83).

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are (5;8) and (1;83).


Submit your answer as: and

Substitution with quadratic equations

Here are two equations, which we can solve simultaneously using substitution:

y=x2+x7y=2x1

Which of the following equations shows the correct substitution step?

A 2x1=y2+y7
B y=(2x)2+(2x)7
C 2y1=x2+x7
D 2x1=x2+x7
Answer: The correct substitution step is choice: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to substitute 2x1 in place of y in the quadratic equation y=x2+x7.


STEP: Substitute the linear equation into the quadratic equation
[−1 point ⇒ 0 / 1 points left]

We have two equations that we will solve by substitution. The equations are:

quadratic equationy=x2+x7linear equationy=2x1

We need to substitute 2x1 from the linear equation in place of y in the quadratic equation.

2x1=x2+x7
NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. The solution for this question did not consider this case.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for x. The question does not ask us to solve the simultaneous equations.

The correct substitution step is 2x1=x2+x7 which is choice D.


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Substitution with quadratic equations

Here are two equations, which we can solve simultaneously using substitution:

y=x2x+5y=2x+7

Which of the following equations shows the correct substitution step?

A 2x+7=x2x+5
B 2x+7=y2y+5
C y=(2x+7)2(2x+7)+5
D 2x+7=x2x
Answer: The correct substitution step is choice: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to substitute 2x+7 in place of y in the quadratic equation y=x2x+5.


STEP: Substitute the linear equation into the quadratic equation
[−1 point ⇒ 0 / 1 points left]

We have two equations that we will solve by substitution. The equations are:

quadratic equationy=x2x+5linear equationy=2x+7

We need to substitute 2x+7 from the linear equation in place of y in the quadratic equation.

2x+7=x2x+5
NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. The solution for this question did not consider this case.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for x. The question does not ask us to solve the simultaneous equations.

The correct substitution step is 2x+7=x2x+5 which is choice A.


Submit your answer as:

Substitution with quadratic equations

Here are two equations, which we can solve simultaneously using substitution:

y=x22x4y=3x8

Which of the following equations shows the correct substitution step?

A y=(3x)22(3x)4
B 3x8=x22x4
C y=(3x8)22(3x8)4
D 3x8=y22y4
Answer: The correct substitution step is choice: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to substitute 3x8 in place of y in the quadratic equation y=x22x4.


STEP: Substitute the linear equation into the quadratic equation
[−1 point ⇒ 0 / 1 points left]

We have two equations that we will solve by substitution. The equations are:

quadratic equationy=x22x4linear equationy=3x8

We need to substitute 3x8 from the linear equation in place of y in the quadratic equation.

3x8=x22x4
NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. The solution for this question did not consider this case.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for x. The question does not ask us to solve the simultaneous equations.

The correct substitution step is 3x8=x22x4 which is choice B.


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Equations with fractions

Solve the following equations simultaneously to find the values of x and y:

4x7xy=4yx,y04y=2x5

Your answer should be exact (do not round off).

Answer:

The solution is x= and y= .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The first thing you should do is rearrange the first equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is xy.


STEP: Rearrange the first equation
[−1 point ⇒ 4 / 5 points left]

We need to solve the equations simultaneously. To do that, we should start by rearranging the first equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by xy, which is the LCD of the fractions. That will cancel all of the denominators.

4x7xy=4yxy(4x7xy)=xy(4y)4xxy7=4yxy4y7=4x

STEP: Pick a method and solve for whichever variable comes more easily
[−3 points ⇒ 1 / 5 points left]

Now we are ready to solve the following equations:

4y7=4x4y=2x5

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations invite elimination, because the y-terms have equal and opposite coefficients. We can eliminate the y-terms by adding the equations.

Eliminate the y and solve for x:

4y7=4xsecond equationAdd the+(4y)=+(2x5)74y+4y=52x+4x7=2x52x=2x=1

Great - we have the first value, x=1.


STEP: Solve for the other variable
[−1 point ⇒ 0 / 5 points left]

Now the last step: find the value of y. We can do this using either of the equations in the question: but the second eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the first equation).

4y=2x54y=2(1)54y=3y=34

The answer is the pair of numbers x=1 and y=34. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

NOTE: Remember, the fractions force x,y0, which is why there are open intervals at the x- and y-intercepts of the fraction equation.

The values which solve these two equations are x=1 and y=34.


Submit your answer as: and

Equations with fractions

Solve the following equations simultaneously to find the values of x and y:

1y=4x9xyx,y02x15=4y

Your answer should be exact (do not round off).

Answer:

The solution is x= and y= .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The first thing you should do is rearrange the first equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is xy.


STEP: Rearrange the first equation
[−1 point ⇒ 4 / 5 points left]

We need to solve the equations simultaneously. To do that, we should start by rearranging the first equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by xy, which is the LCD of the fractions. That will cancel all of the denominators.

1y=4x9xyxy(1y)=xy(4x9xy)xyy=4xxy9x=4y9

STEP: Pick a method and solve for whichever variable comes more easily
[−3 points ⇒ 1 / 5 points left]

Now we are ready to solve the following equations:

x=4y92x15=4y

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations invite elimination, because the y-terms have equal and opposite coefficients. We can eliminate the y-terms by adding the equations.

Eliminate the y and solve for x:

x=4y9second equationAdd the+(2x15)=+(4y)152x+x=94y+4yx15=9x=6x=6

Great - we have the first value, x=6.


STEP: Solve for the other variable
[−1 point ⇒ 0 / 5 points left]

Now the last step: find the value of y. We can do this using either of the equations in the question: but the second eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the first equation).

2x15=4y2(6)15=4y4y=3y=34

The answer is the pair of numbers x=6 and y=34. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

NOTE: Remember, the fractions force x,y0, which is why there are open intervals at the x- and y-intercepts of the fraction equation.

The values which solve these two equations are x=6 and y=34.


Submit your answer as: and

Equations with fractions

What are the values of x and y which solve these equations simultaneously?

6xy=6y1xx,y04x=6y4

Your answer should be exact (do not round off).

Answer:

The solution is x= and y= .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The first thing you should do is rearrange the first equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is xy.


STEP: Rearrange the first equation
[−1 point ⇒ 4 / 5 points left]

We need to solve the equations simultaneously. To do that, we should start by rearranging the first equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by xy, which is the LCD of the fractions. That will cancel all of the denominators.

6xy=6y1xxy(6xy)=xy(6y1x)6=1xxy+6yxy6=6xy

STEP: Pick a method and solve for whichever variable comes more easily
[−3 points ⇒ 1 / 5 points left]

Now we are ready to solve the following equations:

6=6xy4x=6y4

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations are not arranged nicely for elimination or substitution. We must change the equations somehow. In this case, we can make a small change that will allow us to use elimination: we can multiply the first equation by 6 so that the y-coefficients will be ready to cancel. Then we can eliminate them by adding the equations.

First modify things to set up the elimination:

6(6)=6(6xy)36=36x6y

Now do the elimination and complete the solution to find x.

36=36x6ysecond equationAdd the+(4x)=+(6y4)36+4x=4+36x6y+6y4x+36=36x432x=40x=54

Great - we have the first value, x=54.


STEP: Solve for the other variable
[−1 point ⇒ 0 / 5 points left]

Now the last step: find the value of y. We can do this using either of the equations in the question: but the second eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the first equation).

4x=6y44(54)=6y46y=9y=32

The answer is the pair of numbers x=54 and y=32. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

NOTE: Remember, the fractions force x,y0, which is why there are open intervals at the x- and y-intercepts of the fraction equation.

The values which solve these two equations are x=54 and y=32.


Submit your answer as: and

Simultaneous equations: special outcomes

Solve for x and y:

y=3x25x+1y=x2
INSTRUCTION: You should enter your answer in the form of a coordinate pair, (x; y). An example of an acceptable answer is (2; 5). If there is no solution, type no solution.
Answer: The solution is .
coordinate
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to start by substituting x2 in place of y in the quadratic equation and simplify. You must expect a special kind of solution.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 5 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is y=3x25x+1 and the linear equation is y=x2. This means we expect two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
x2=3x25x+10=3x25x+1x+20=3x25xx+1+20=3x26x+3

STEP: Solve the equation 0=3x26x+3
[−2 points ⇒ 1 / 5 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE:We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

We will factorise using the grouping method. For this method, we need two numbers that add to give 6. This is the coefficient of the middle term. The product of the two numbers must be 9. This is the product of the coefficient of the first term 3 and the value of the last term 3. If we take 3 which is half of the value of 6 (the coefficient of the middle term) and square it, we get the product 9. In this case we will use only one number to factorise, that is, 3. We will re-write the middle term, forming two terms, using this value.

0=3x26x+30=3x23x3x+3

We now group the first two terms together and group the last terms together and factorise.

0=3x23x3x+30=3x(x1)3(x1)

The last thing now is to factor the highest common factor.

0=(3x3)(x1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. In this case we have one root, 1. This root is the x-value of our solution. This x-value will have one corresponding y value.


STEP: Calculate the corresponding value of y
[−1 point ⇒ 0 / 5 points left]

Now that we have the value of x, we will substitute it into the linear equation to calculate the corresponding value of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.
y=x2=1(1)2=1

The solution is x=1 together with y=1. Using coordinate pairs, the solution is (1;1).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (1;1). This is the same solution we have calculated. This means we are 100% correct.

The solution is (1;1).


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Simultaneous equations: special outcomes

Solve for x and y:

y=3x2+5x+4y=2x3
INSTRUCTION: You should enter your answer in the form of a coordinate pair, (x; y). An example of an acceptable answer is (2; 5). If there is no solution, type no solution.
Answer: The solution is .
one-of
type(string.nocase)
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to start by substituting 2x3 in place of y in the quadratic equation and simplify. You must expect a special kind of solution.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 2 / 4 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is y=3x2+5x+4 and the linear equation is y=2x3. This means we expect two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
2x3=3x2+5x+40=3x2+5x+4+2x+30=3x2+5x+2x+4+30=3x2+7x+7

STEP: Solve the equation 0=3x2+7x+7
[−2 points ⇒ 0 / 4 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

We need two numbers that add to give 7. This is the coefficient of the middle term. The product of the two numbers must be 21. This is the product of the coefficient of the first term 3 and the value of the last term 7. In this case, there are no such numbers. We will now use the quadratic formula to solve this equation.

The solution for a quadratic equation of the form ax2+bx+c=0:

x=b±(b24ac)2a

Before we can use the quadratic formula, we need to identify the coefficients a, b and c from the equation 3x2+7x+7=0 These are:

a=3,b=7 and c=7.

We will substitute these values into the formula and simplify.

x=(7)±((7)24(3)(7))2(3)=7±(49(84))6=7±(35)6

The number under the square root sign is negative. This means that our quadratic equation has no real solution. This means that y=3x2+5x+4 and y=2x3 cannot be solved simultaneously.

We can confirm our result by plotting the graph of each equation on the same set of axes. We do not expect these graphs to intersect.

Aha! The two graphs do not have an intersection point. This means that indeed the two equations have no solution.

The final answer is no solution.


Submit your answer as:

Simultaneous equations: special outcomes

Solve for x and y from the given equations:

y=x22x+2y=x5
INSTRUCTION: You should enter your answer in the form of a coordinate pair, (x; y). An example of an acceptable answer is (2; 5). If there is no solution, type no solution.
Answer: The solution is .
one-of
type(string.nocase)
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to start by substituting x5 in place of y in the quadratic equation and simplify. You must expect a special kind of solution.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 2 / 4 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is y=x22x+2 and the linear equation is y=x5. This means we expect two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
x5=x22x+20=x22x+2x+50=x22xx+2+50=x23x+7

STEP: Solve the equation 0=x23x+7
[−2 points ⇒ 0 / 4 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

We need two numbers that add to give 3. This is the coefficient of the middle term. The product of the two numbers must be 7. This is the product of the coefficient of the first term 1 and the value of the last term 7. In this case, there are no such numbers. We will now use the quadratic formula to solve this equation.

The solution for a quadratic equation of the form ax2+bx+c=0:

x=b±(b24ac)2a

Before we can use the quadratic formula, we need to identify the coefficients a, b and c from the equation x23x+7=0 These are:

a=1,b=3 and c=7.

We will substitute these values into the formula and simplify.

x=(3)±((3)24(1)(7))2(1)=3±(9(28))2=3±(19)2

The number under the square root sign is negative. This means that our quadratic equation has no real solution. This means that y=x22x+2 and y=x5 cannot be solved simultaneously.

We can confirm our result by plotting the graph of each equation on the same set of axes. We do not expect these graphs to intersect.

Aha! The two graphs do not have an intersection point. This means that indeed the two equations have no solution.

The final answer is no solution.


Submit your answer as:

Simultaneous equations

Adapted from DBE Nov 2016 Grade 12, P1, Q1.3
Maths formulas

Solve for x and y:

x=4yandx2+2xy=32
INSTRUCTIONS:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are
and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.


STEP: Substitute x=4y into x2+2xy=32 and solve for y
[−3 points ⇒ 1 / 4 points left]

We need to solve the following pair of equations for x and y:

x=4yandx2+2xy=32

First, we can substitute 4y in place of x in the second equation:

(4y)2+2(4y)(y)=32

We can simplify this equation to solve for y:

(4y)2+2(4y)(y)=3216y28y2=328y2=32y2=4

y can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about y2. So we have to consider both the positive and the negative values as solutions to the equations.

y=2 or y=2

STEP: Use the y-values to find the x-values
[−1 point ⇒ 0 / 4 points left]

Each of the y-values that we have found will have a corresponding value of x, which can be found from the equation x=4y.

For y=2:

x=4y=4(2)=8

And for y=2:

x=4y=4(2)=8

So the x-values are x=8 or x=8.

Therefore the correct coordinate pairs are (8;2) and (8;2).


Submit your answer as: and

Simultaneous equations

Adapted from DBE Nov 2016 Grade 12, P1, Q1.3
Maths formulas

Solve for x and y:

x=2yandx24xy=16
INSTRUCTIONS:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are
and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.


STEP: Substitute x=2y into x24xy=16 and solve for y
[−3 points ⇒ 1 / 4 points left]

We need to solve the following pair of equations for x and y:

x=2yandx24xy=16

First, we can substitute 2y in place of x in the second equation:

(2y)24(2y)(y)=16

We can simplify this equation to solve for y:

(2y)24(2y)(y)=164y28y2=164y2=16y2=4

y can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about y2. So we have to consider both the positive and the negative values as solutions to the equations.

y=2 or y=2

STEP: Use the y-values to find the x-values
[−1 point ⇒ 0 / 4 points left]

Each of the y-values that we have found will have a corresponding value of x, which can be found from the equation x=2y.

For y=2:

x=2y=2(2)=4

And for y=2:

x=2y=2(2)=4

So the x-values are x=4 or x=4.

Therefore the correct coordinate pairs are (4;2) and (4;2).


Submit your answer as: and

Simultaneous equations

Adapted from DBE Nov 2016 Grade 12, P1, Q1.3
Maths formulas

Solve for x and y:

x=2yandx2+4xy=36
INSTRUCTIONS:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are
and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.


STEP: Substitute x=2y into x2+4xy=36 and solve for y
[−3 points ⇒ 1 / 4 points left]

We need to solve the following pair of equations for x and y:

x=2yandx2+4xy=36

First, we can substitute 2y in place of x in the second equation:

(2y)2+4(2y)(y)=36

We can simplify this equation to solve for y:

(2y)2+4(2y)(y)=364y28y2=364y2=36y2=9

y can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about y2. So we have to consider both the positive and the negative values as solutions to the equations.

y=3 or y=3

STEP: Use the y-values to find the x-values
[−1 point ⇒ 0 / 4 points left]

Each of the y-values that we have found will have a corresponding value of x, which can be found from the equation x=2y.

For y=3:

x=2y=2(3)=6

And for y=3:

x=2y=2(3)=6

So the x-values are x=6 or x=6.

Therefore the correct coordinate pairs are (6;3) and (6;3).


Submit your answer as: and

Simultaneous equations with xy terms

Determine the solution of the following equations:

2y=4xxy2y=2x+4
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting 2x+4 in place of y in the hyperbolic equation and then solve for x.


STEP: Substitute the linear equation into the hyperbolic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

hyperbolic equation2y=4xxy2linear equationy=2x+4

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

NOTE: We could also isolate either x or y from 2y=4xxy2 and substitute its value into y=2x+4. We could still get the same answer, but the calculations might become too long.
2(2x+4)=4xx(2x+4)20=4xx(2x+4)2+2(2x+4)0=4x(2x2+4x)2+(4x+8)0=2x28x+4x2+80=2x24x2+80=2x24x+6make the first term positivedivide each term by -1 to 0=2x2+4x6

STEP: Solve the equation 0=2x2+4x6
[−2 points ⇒ 2 / 6 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=2x2+4x60=(x+3)(x1)

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 3 and 1.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

By using the values of x, we will calculate the corresponding values of y. We will do so by substituting each x-value into the linear equation.

NOTE: We can still get the same values of y if we substitute the values x into the equation 2y=4xxy2. If we do that, the calculations might become too long.

For x=3:

y=2x+4=2

For x=1:

y=2x+4=6

The solutions are x=3 together with y=2 , and x=1 with y=6. Using coordinate pairs, the solutions are (3;2) and (1;6).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (3;2) and (1;6). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (3;2) and (1;6).


Submit your answer as: and

Simultaneous equations with xy terms

Solve for x and y from the given equations:

y=2xxy+1x=3y+5
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting 3y+5 in place of x in the hyperbolic equation and then solve for y.


STEP: Substitute the linear equation into the hyperbolic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

hyperbolic equationy=2xxy+1linear equationx=3y+5

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of x from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

NOTE: We could also isolate either x or y from y=2xxy+1 and substitute its value into x=3y+5. We could still get the same answer, but the calculations might become too long.
y=2(3y+5)(3y+5)y+10=2(3y+5)(3y+5)y+1y0=2(3y+5)y(3y+5)+1y0=3y26y5yy10+10=3y211yy10+10=3y212y10+10=3y212y9make the first term positivedivide each term by -1 to 0=3y2+12y+9

STEP: Solve the equation 0=3y2+12y+9
[−2 points ⇒ 2 / 6 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=3y2+12y+90=(y+1)(y+3)

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are 1 and 3.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

By using the values of y, we will calculate the corresponding values of x. We will do so by substituting each y-value into the linear equation.

NOTE: We can still get the same values of x if we substitute the values y into the equation y=2xxy+1. If we do that, the calculations might become too long.

For y=1:

x=3y+5=2

For y=3:

x=3y+5=4

The solutions are x=2 together with y=1 , and x=4 with y=3. Using coordinate pairs, the solutions are (2;1) and (4;3).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (2;1) and (4;3). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (2;1) and (4;3).


Submit your answer as: and

Simultaneous equations with xy terms

Solve for x and y:

2y=6xxy+2y=2x10
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting 2x10 in place of y in the hyperbolic equation and then solve for x.


STEP: Substitute the linear equation into the hyperbolic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

hyperbolic equation2y=6xxy+2linear equationy=2x10

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

NOTE: We could also isolate either x or y from 2y=6xxy+2 and substitute its value into y=2x10. We could still get the same answer, but the calculations might become too long.
2(2x10)=6xx(2x10)+20=6xx(2x10)+2+2(2x10)0=6x(2x210x)+2+(4x20)0=2x2+16x+4x+2200=2x2+20x+2200=2x2+20x18make the first term positivedivide each term by -1 to 0=2x220x+18

STEP: Solve the equation 0=2x220x+18
[−2 points ⇒ 2 / 6 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=2x220x+180=(x1)(x9)

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 1 and 9.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

By using the values of x, we will calculate the corresponding values of y. We will do so by substituting each x-value into the linear equation.

NOTE: We can still get the same values of y if we substitute the values x into the equation 2y=6xxy+2. If we do that, the calculations might become too long.

For x=1:

y=2x10=8

For x=9:

y=2x10=8

The solutions are x=1 together with y=8 , and x=9 with y=8. Using coordinate pairs, the solutions are (1;8) and (9;8).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (1;8) and (9;8). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (1;8) and (9;8).


Submit your answer as: and

3. Practical applications

Setting up simultaneous equations

Here are facts about two numbers:

  • The sum of the numbers is 13.
  • The numbers are consecutive numbers.

To find these numbers, we can write equations for each of these facts. Let n1 represent one of the numbers and n2 represent the other number. Then which equations accurately represent each fact? Select your answer from the choices below.

Answer:
Fact about the numbers Equation
The sum of the numbers is 13.
The numbers are consecutive numbers.
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let n1 represent one of the numbers and n2 represent the other number". So we can break up the first fact like this:

The sum of the numbersis13n1+n2=13

The correct equation for the first fact is n1+n2=13.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like 10 and 11, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is 1 more than the small number." Another is "the small number is 1 less than the larger number." Let's use the second option.

The smaller numberis1 less than the larger numbern1=n21

This makes sense: consecutive numbers are separated by 1. The equation shows that with the 1.

The correct answers are:

Fact about the numbers Equation
The sum of the numbers is 13. n1+n2=13
The numbers are consecutive numbers. n1=n21

Submit your answer as: and

Word problems: distance, speed and time

Two ships start out at two different ports. The ships are 141 km apart. The ships start moving towards each other. One ship is moving at 72 km/h and the other ship at 69 km/h.

If both ships started their journey at the same time, how long will they take to pass each other?

Answer: The two ships pass each other after hours.
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.


STEP: Sketch a simple picture to organise the information
[−2 points ⇒ 4 / 6 points left]

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two ships must be equal to the total distance when the ships meet:

d1+d2=dtotald1+d2=141 km

STEP: Write equations to describe the motion of the ships
[−2 points ⇒ 2 / 6 points left]

This question is about distances, speeds, and times. The equation connecting these values is

speed =distance time OR distance =speed ×time

We want to know the amount of time needed for the ships to meet. Let the time taken be t. Then we can write an expression for the distance each of the ships travels:

For ship 1:

d1=s1tship is 72 km/hThe speed of the firstd1=72t

For ship 2:

d2=s2tship is 69 km/hThe speed of the secondd2=69t

STEP: Solve the equations simultaneously for t
[−2 points ⇒ 0 / 6 points left]

Now we have three different equations. We can combine them using substitution to solve for the value of t.

d1+d2=141(72t)+(69t)=141141t=141t=141141t=1

The ships will meet after 1 hours.


Submit your answer as:

Word problems: solving simultaneous equations

A group of friends is buying lunch together. The group buys 2 wraps and 4 hamburgers. Here are some facts about their lunch:

  • the total cost for the 2 wraps and 4 hamburgers is R166
  • a wrap costs R8 more than a hamburger

What is the price for one wrap and the price for one hamburger?

Answer:

A wrap costs R and a hamburger costs R .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

You need to choose variables to represent the things you want to find and write equations based on the information in the question.


STEP: Pick variables for the things we want to know
[−1 point ⇒ 6 / 7 points left]

The two things we want to know in this question are the prices for each of the items (a wrap and a hamburger). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

w=the price of a wraph=the price of a hamburger

STEP: Write equations based on the information in the question
[−2 points ⇒ 4 / 7 points left]

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the 2 wraps and 4 hamburgers is R166." We can use the expression 2w to represent the price of the 2 wraps. Similarly, the expression 4h represents the price of the 4 hamburgers. With these values we can write a full equation for the prices:

In words: the total cost for the 2 wraps and 4 hamburgers is R166In maths: 2w+4h=166

Now we can use the second point. It says, "a wrap costs R8 more than a hamburger." We can write this as an equation like this:

In words: a wrap costs R8 more than a hamburgerIn maths: w=h+8

STEP: Solve the equations simultaneously
[−2 points ⇒ 2 / 7 points left]

Now that we have two equations, we need to solve them simultaneously.

2w+4h=166w=h+8

We could use elimination, but substitution is a better choice (because w is already the subject). Substitute the second equation into the first equation and solve!

2w+4h=1662(h+8)+4h=1662h+16+4h=1666h=16616h=1506=25

This means that the price of one hamburger is R25.


STEP: Find the other variable's value
[−1 point ⇒ 1 / 7 points left]

Finally, use the value we found for h to find the value of w (the price of a wrap). We can use either equation to do this, but the second one is easier to use (because w is already isolated).

w=h+8=25+8=33

The price for one wrap is R33.


STEP: Write the final answer
[−1 point ⇒ 0 / 7 points left]

It is important to write the answer to a word problem as a complete sentence.

The price of the wrap is R33 while a hamburger costs R25.


Submit your answer as: and

Word problems: checking answers about shopping

Suppose you must solve this word problem:

At a shop there are some fruits and sweets for sale. Ben buys some bananas and some sweets. The total cost for 5 bananas and 5 sweets is R32,50. And each sweet costs R0,30 less than each banana. What is the price for one banana?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the price be a decimal number (or must it be an integer)?
  2. Can the price be a negative number?
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the price of each banana. Can the price be a decimal, like 4,5? Can it be negative, like 5?


STEP: Decide if the price can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be R6,75. Or R5. Or R19,87. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the price can be a decimal number or not. In other words, can the price of something be a decimal number like R4,30, or must a price always be a whole number like R4,00? Of course we know that many prices are not whole numbers! This is similar to the example above about airtime, which can be a decimal number. There is no reason why the price of an item must be a whole number.

The answer for the first question is: Yes, the price can be a decimal number.


STEP: Decide if the price can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if the price can be negative, or must be positive. Prices cannot be negative. Imagine something which costs -R5,50. That is impossible!

The answer to the second question is: No, the price cannot be a negative number.

The correct answer choices are:

  1. Yes, it can be a decimal
  2. No, it cannot be negative

Submit your answer as: and

Word problems: checking answers with shapes

Suppose you must solve this word problem:

Yaseen is drawing a pattern of rectangles in his notebook. The number of rectangles in each figure is 2 more than the number of rectangles in the previous figure. If there are 10 rectangles in the fourth figure, how many rectangles are there in the first figure?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the number of rectangles be a decimal number (or must it be an integer)?
  2. Can the number of rectangles be a negative number?
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the number of rectangles in the first figure of Yaseen's diagram. Can the number of rectangles be a decimal, like 4,5? Can it be negative, like 5?


STEP: Decide if the number of rectangles can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be 150 days of school, shame. Similarly, it is not possible to have 176,3562 days of school. What type of numbers are realistic for the number of rectangles?

For the first question we must decide if the number of rectangles can be a decimal number. Is that fine or is it impossible?

In other words, can the number of rectangles be a decimal number like 3,7, or must the number be a whole number like 4? Just like the number of days in the school year, the number of rectangles cannot be a decimal value. 3,7 rectangles does not make sense! The number of rectangles must be a whole number! (You might think, "That's wrong, we can just draw part of a rectangle to get 3,7 rectangles." But if the shape is not complete, it is not a rectangle.)

The answer for the first question is: No, it must be an integer, the number of rectangles cannot be a decimal number.


STEP: Decide if the number of rectangles can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if number of rectangles can be negative, or must be positive. The number of rectangles cannot be negative. If Yaseen drew any rectangles at all, then the answer to this question must be positive. There is no such thing as 4 rectangles.

The answer for the second question is: No, the number of rectangles cannot be a negative number.

The correct answer choices are:

  1. No, it must be an integer
  2. No, it cannot be negative

Submit your answer as: and

Word problems: test scores and simultaneous equations

Hassana and Chibueze are friends. Hassana takes Chibueze's consumer studies test paper and says: “I have 20 marks more than you do and the sum of both our marks is equal to 134. What are our marks?”

Answer:

Hassana got marks and Chibueze got marks for the consumer studies test paper.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.


STEP: Pick variables for the marks
[−1 point ⇒ 5 / 6 points left]

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

h= Hassana's markc= Chibueze's mark

STEP: Write equations about the students' marks
[−2 points ⇒ 3 / 6 points left]

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

 more than ChibuezeHassana has 20 marksh=c+20(the total) is 134The sum of the marksh+c=134

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

h+c=134(c+20)+c=1342c=13420c=1142=57

This means that Chibueze's mark is 57.


STEP: Find Hassana's mark
[−1 point ⇒ 0 / 6 points left]

Now we can use Chibueze's mark and one of the equations we have to find Hassana's mark. The easiest way to do that is to substitute Chibueze's mark back into the first equation:

h=c+20=(57)+20=77

The students achieved these marks: Hassana earned 77 and Chibueze earned 57.


Submit your answer as: and

Word problems: odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive odd integers.
  • The sum of the numbers is 24.

Determine the value of the larger number.

Answer: The larger number is .
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. For example, you can use n1 and n2. Then write equations using these variables based on the facts given in the question.
STEP: Pick variables
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find one of them (the larger one). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a sum of 24. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

n1=the smaller numberwe need to findthis is the numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact given in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, 11 and 13 are consecutive odd numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the even integer which sits between n1 and n2.

The second fact about the numbers tells us that "the sum of the numbers is 24". Remember that sum means addition. So:

n1+n2=24

STEP: Solve the equations for n2
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the larger number (we do not need both of them). That means we need to find the value of n2.

We can solve this using substitution. However, remember that we want the value of n2 (the larger number). We can start by rearranging the first equation to make n1 the subject. Then the substitution step will remove n1 from the second equation and we can solve for n2. (This is not required - it just makes the solution faster.)

n2=n1+2n22=n1

Now substitute n22 into the other equation and solve for n2.

(n22)+n2=242n22=242n2=26n2=13

The result is n2=13. Notice that this means that the other number, n1, must be 11, because n2=n1+2. This is perfect, because we know that the sum of the numbers is 24, and 13+11=24.

The larger number is 13.


Submit your answer as:

Word problems: checking answers

Emma is 12 years younger than her brother, Songezo. In 10 years, Songezo will be 2 times as old as Emma. How old is Emma now?

INSTRUCTION: If there is no acceptable solution, type 'no solution' in the answer box.
Answer:

Emma is years old.

numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to find Emma's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.


STEP: Choose variables for each person's age
[−1 point ⇒ 4 / 5 points left]

We need to determine Emma's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Songezo and Emma. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

aS=Songezo's ageaE=Emma's age

STEP: Write an equation based on the information in the question
[−1 point ⇒ 3 / 5 points left]

Now let's write equations using these variables. From the question we know that: "Emma is 12 years younger than her brother, Songezo." We need to translate that into mathematics. The key word is younger, which tells us to use subtraction to relate the ages.

aE=aS12

STEP: Write the equations
[−1 point ⇒ 2 / 5 points left]

We also know that: "In 10 years, Songezo will be 2 times as old as Emma." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

aS+10=Songezo's age in 10 yearsaE+10=Emma's age in 10 years

These are the ages at which Songezo will be 2 times as old as Emma. We can put all this together as follows:

in 10 yearsSongezo's age=2×in 10 yearsEmma's ageaS+10=2(aE+10)

STEP: Solve the equations simultaneously
[−2 points ⇒ 0 / 5 points left]

Now we have to solve two equations, both of them including the variables aS and aE. The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is aE=aS12. We can substitute this into the second equation and solve for aS.

aS+10=2(aE+10)aS+10=2((aS12)+10)aS+10=2aS414=aS

Terrific: this means that Songezo is 14 years old. But the question asked for us to find Emma's age. We can find it using the first equation, which relates the two ages:

aE=aS12=(14)12=2

So we finally got the answer to the question. Emma is 2 years old.

The correct answer is: 2.


Submit your answer as:

Word problems: products of odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive even integers.
  • The product of the numbers is 48.

Find the value of the smaller number.

Answer: The smaller number is .
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. Then write equations using these variables based on the information given in the question.
STEP: Pick variables for the numbers
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find the smaller number (not both). We know certain things about these numbers: they are positive, they are consecutive even integers, and they have a product of 48. We can use that information to solve this question using simultaneous equations.

The first thing we need to do is define variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent the things we want to find. In this case we are looking for two numbers, so these are good choices:

we need to findthis is the numbern1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact in the question says that the numbers are "consecutive even integers". So both of the numbers are even, and they come one after another. For example, 10 and 12 are consecutive even numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the odd integer which sits between n1 and n2.

The second fact about the numbers tells us that "the product of the numbers is 48". Remember that product means multiplication. So:

n1n2=48

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

If we subsitute n2=n1+2 into the equation n1n2=48, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1(n1+2)=48n12+2n1=48n12+2n148=0(n1+8)(n16)=0
n1=8andn1=6

This solution led to two answers for n1. However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves n1=6.

This means the other number, n2, must be 8, because n2=n1+2. This is perfect, because we also know that the product of the numbers is 48, and 68=48.

The smaller number is 6.


Submit your answer as:

Word problems: finding consecutive numbers

Here are some facts about two numbers:

  • the numbers are consecutive integers
  • both numbers are positive
  • the product of the numbers is 210

What are the two numbers?

TIP: 'Product' means multiplication; 'consecutive' means 'following each other without gaps between.'
INSTRUCTION: Write your answers in the boxes below. It does not matter what order your answers are in.
Answer: The numbers are and .
numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You should represent the numbers in the question by n and n+1. Write an equation based on the fact that the product of those two numbers is 210.


STEP: Pick variables to represent the numbers
[−1 point ⇒ 5 / 6 points left]

This question asks us to find two numbers which have a product of 210. That means if we multiply the numbers we get 210.

The first thing we need to do is pick variables for the numbers we are trying to find. We can pick any variable. But it is a good idea to pick a variable which is related to whatever it represents. In this question, n is a good choice because we are looking for numbers. In fact, since there are two different numbers, we can do this:

n1=the first numbern2=the next number

STEP: Write an equation based on the information in the question
[−1 point ⇒ 4 / 6 points left]

The question tells us that the product of these numbers is 210. In other words, if we multiply the numbers, we get 210. As an equation this is:

n1n2=210

STEP: Connect the two variables
[−1 point ⇒ 3 / 6 points left]

Now we can use the fact that the numbers are consecutive. So if n1 represents the first number, then the next number must be one more than n1.

If:n2=the number following n1then:n2=n1+1

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have two equations about the two numbers.

n1n2=210n2=n1+1

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for n1.

n1(n1+1)=210(n1)2+n1210=0(n1+15)(n114)=0
n1=15andn1=14

The solutions to the equation are n1=15 or n1=14. This means the first number is either −15 or 14.


STEP: Find the final answers
[−1 point ⇒ 0 / 6 points left]

Wait a minute! The numbers −15 and 14 are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are positive. That means we must throw out the n1=15, so the first number is n1=14.

Based on that we can find the second number. The second number is n1+1, which is equal to (14)+1=15.

The two consecutive integers are 14 and 15.


Submit your answer as: and

Word problems: an age-old question

Talwar has a son, Roy. Here are some facts about how old Talwar and Roy are:

  • Talwar is 9 times as old as Roy right now.
  • 9 years from now, Talwar will be 3 times as old as Roy.

How old are Talwar and Roy now?

Answer:

Talwar is years old and Roy is years old.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.


STEP: Choose variables for the information
[−1 point ⇒ 5 / 6 points left]

In this question we want to find the age of two people: Talwar and his son, Roy. We can solve this by setting up two equations and solving them simultaneously.

Start by choosing variables to represent the ages of the father and the son. (We need to do this because we don't know the ages!) It is a good idea to choose variables that match what we are describing. So:

Let t=Talwar's ageLet r=Roy's age

STEP: Write equations based on the information in the question
[−2 points ⇒ 3 / 6 points left]

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Talwar is now 9 times as old as Roy." As an equation, this is:

Ages now: t=9r

The second piece of information says that in "9 years... Talwar will be 3 times as old as his son." In 9 years Talwar will be t+9 years old, and similarly Roy will be r+9 years old. Then:

Ages in 9 years: t+9=3(r+9)

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have a pair of simultaneous equations! Substitute the first equation into the second equation and solve. (You can solve the equations using elimination if you prefer. In that case, the easiest option is to subtract the equations to cancel t.)

t+9=3(r+9)(9r)+9=3r+276r=18r=3

Great! Now we know that Roy is 3 years old.


STEP: Use Roy's age to find Talwar's age
[−1 point ⇒ 0 / 6 points left]

We can now find Talwar's age. Substitute Roy's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

t=9r=9(3)=27

Write your final answer: Talwar is 27 years old and Roy is 3 years old.


Submit your answer as: and

Word Problems: checking answers

Suppose you must solve this word problem:

A marathon is a long-distance running race with an official distance of 42,195 kilometres. 1 000 people ran the "Sunshine and Roses are for Sissies" marathon in Durban. The second-place runner finished 12,72 seconds after the winner. The third runner was 13,55 seconds behind the second runner, and the fourth runner was 26,15 seconds behind the third runner. If the second-place runner had an average speed of 19,15 km/h, how many of these runners finished the marathon in less than 2,24 hours?

Can the answer to the question above be a negative number?

Answer:

Can the answer be negative?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Think about the meaning of the number you need. Some values cannot be decimal numbers. Other values cannot be negative numbers. For example, a distance can be a decimal, but it cannot be negative.


STEP: Decide if the answer can be negative
[−1 point ⇒ 0 / 1 points left]

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a negative number or not.

Focus on what the problem asks: how many of these runners finished the marathon in less than 2,24 hours. So the question is about the number of people who finished the race before 2,24 hours had passed. The answer cannot be negative. The number of people to finish before 2,24 hours might be 3 or 7 people, but it cannot be 5 people. In other words, the number of people must be a positive number or zero, but it cannot be negative.

The correct answer is: No, it cannot be negative.


Submit your answer as:

Word problems: rectangle facts

The diagonal of a rectangle is 16 cm more than its width. The length of the same rectangle is 8 cm more than its width.

Determine the width and length of the rectangle.

Answer: The width is cm and length is cm.
numeric
numeric
HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.


STEP: Draw a diagram and label it
[−1 point ⇒ 7 / 8 points left]

Drawing a diagram is a helpful way to organise the information in a question about a shape. This question is about a rectangle, so we can start with a rectangle picture. The question mentions the diagonal of the ractangle, so we should draw that also.

Notice we have added labels for the three parts of the rectangle mentioned in the question: d is the diagonal, w is the width, and l is the length.


STEP: Connect the variables in the diagram
[−2 points ⇒ 5 / 8 points left]

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

16 cm more than its widththe diagonal of the rectangle isd=w+168 cm more than its widththe length of the rectangle isl=w+8

STEP: Use the Theorem of Pythagoras and solve the equations simultaneously
[−2 points ⇒ 3 / 8 points left]

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

From the picture we can see that there are two right-angled triangles in the rectangle. As always for a right-angled triangle, we can use the Theorem of Pythagoras. Using our variables from the diagram, we can write:

d2=w2+l2

And now we can substitute in the equations from above. Specifically we can substitute w+16 in for d and w+8 in for l. Then expand the binomials and simplify the equation as much as possible.

d2=w2+l2(w+16)2=w2+(w+8)2w2+32w+256=w2+(w2+16w+64)0=w216w192

STEP: Solve the equation for w
[−2 points ⇒ 1 / 8 points left]

We have a quadratic equation in standard form! It is time to solve it.

0=w216w1920=(w24)(w+8)w=24or w=8

This means that the width of the rectangle is either 24 cm or 8 cm. But the dimensions of a rectangle cannot be negative, so w=24 cm.


STEP: Calculate the length of the rectangle
[−1 point ⇒ 0 / 8 points left]

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects l to w.

l=w+8=(24)+8=32 cm

The width of the rectangle is 24 cm and the length is 32 cm.


Submit your answer as: and

Setting up simultaneous equations

Last week, Anthea and Bokamoso had a physics test. Now they are comparing their marks and they notice these facts:

  • The sum of the marks is 160.
  • Anthea's mark is 8 less than Bokamoso's mark.

Let a represent Anthea's mark and b represent Bokamoso's mark. Then which equations below accurately represent the facts? Select your answer from the choices below.

Answer:
Fact about the test scores Equation
The sum of the marks is 160.
Anthea's mark is 8 less than Bokamoso's mark.
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is less. Use these key words to change the words into operations, for example, addition or multiplication.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable a for Anthea's mark and b for Bokamoso's mark. So we can break up the first fact like this:

The sum of the marksis160a+b=160

The first fact is equivalent to this equation: a+b=160.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

Similarly, we can identify key parts of the second fact, and translate each into an expression.

Anthea's markis8 less than Bokamoso's marka=b8

This equation, a=b8, means that Anthea's mark is less that Bokamoso's mark.

The correct answers are:

Fact about the test scores Equation
The sum of the marks is 160. a+b=160
Anthea's mark is 8 less than Bokamoso's mark. a=b8

Submit your answer as: and

Exercises

Setting up simultaneous equations

Here are facts about two numbers:

  • The sum of the numbers is 15.
  • The numbers are consecutive numbers.

To find these numbers, we can write equations for each of these facts. Let n1 represent one of the numbers and n2 represent the other number. Then which equations accurately represent each fact? Select your answer from the choices below.

Answer:
Fact about the numbers Equation
The sum of the numbers is 15.
The numbers are consecutive numbers.
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let n1 represent one of the numbers and n2 represent the other number". So we can break up the first fact like this:

The sum of the numbersis15n1+n2=15

The correct equation for the first fact is n1+n2=15.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like 10 and 11, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is 1 more than the small number." Another is "the small number is 1 less than the larger number." Let's use the first option.

The larger numberis1 more than the smaller numbern2=n1+1

This makes sense: consecutive numbers are separated by 1. The equation shows that with the +1.

The correct answers are:

Fact about the numbers Equation
The sum of the numbers is 15. n1+n2=15
The numbers are consecutive numbers. n2=n1+1

Submit your answer as: and

Setting up simultaneous equations

Here are facts about two numbers:

  • The product of the numbers is 72.
  • The numbers are consecutive numbers.

To find these numbers, we can write equations for each of these facts. Let n1 represent one of the numbers and n2 represent the other number. Then which equations accurately represent each fact? Select your answer from the choices below.

Answer:
Fact about the numbers Equation
The product of the numbers is 72.
The numbers are consecutive numbers.
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is product. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let n1 represent one of the numbers and n2 represent the other number". So we can break up the first fact like this:

The product of the numbersis72n1×n2=72

The correct equation for the first fact is n1×n2=72.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like 10 and 11, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is 1 more than the small number." Another is "the small number is 1 less than the larger number." Let's use the second option.

The smaller numberis1 less than the larger numbern1=n21

This makes sense: consecutive numbers are separated by 1. The equation shows that with the 1.

The correct answers are:

Fact about the numbers Equation
The product of the numbers is 72. n1×n2=72
The numbers are consecutive numbers. n1=n21

Submit your answer as: and

Setting up simultaneous equations

Here are facts about two numbers:

  • The product of the numbers is 30.
  • The numbers are consecutive numbers.

To find these numbers, we can write equations for each of these facts. Let n1 represent one of the numbers and n2 represent the other number. Then which equations accurately represent each fact? Select your answer from the choices below.

Answer:
Fact about the numbers Equation
The product of the numbers is 30.
The numbers are consecutive numbers.
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is product. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let n1 represent one of the numbers and n2 represent the other number". So we can break up the first fact like this:

The product of the numbersis30n1×n2=30

The correct equation for the first fact is n1×n2=30.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like 10 and 11, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is 1 more than the small number." Another is "the small number is 1 less than the larger number." Let's use the first option.

The larger numberis1 more than the smaller numbern2=n1+1

This makes sense: consecutive numbers are separated by 1. The equation shows that with the +1.

The correct answers are:

Fact about the numbers Equation
The product of the numbers is 30. n1×n2=30
The numbers are consecutive numbers. n2=n1+1

Submit your answer as: and

Word problems: distance, speed and time

Two missiles start out at two different launchers. The missiles are 1 232 km apart. The missiles start moving towards each other. One missile is moving at 291 km/h and the other missile at 325 km/h.

If both missiles started their journey at the same time, how long will they take to pass each other?

Answer: The two missiles pass each other after hours.
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.


STEP: Sketch a simple picture to organise the information
[−2 points ⇒ 4 / 6 points left]

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two missiles must be equal to the total distance when the missiles meet:

d1+d2=dtotald1+d2=1 232 km

STEP: Write equations to describe the motion of the missiles
[−2 points ⇒ 2 / 6 points left]

This question is about distances, speeds, and times. The equation connecting these values is

speed =distance time OR distance =speed ×time

We want to know the amount of time needed for the missiles to meet. Let the time taken be t. Then we can write an expression for the distance each of the missiles travels:

For missile 1:

d1=s1tmissile is 291 km/hThe speed of the firstd1=291t

For missile 2:

d2=s2tmissile is 325 km/hThe speed of the secondd2=325t

STEP: Solve the equations simultaneously for t
[−2 points ⇒ 0 / 6 points left]

Now we have three different equations. We can combine them using substitution to solve for the value of t.

d1+d2=1 232(291t)+(325t)=1 232616t=1 232t=1 232616t=2

The missiles will meet after 2 hours.


Submit your answer as:

Word problems: distance, speed and time

Two planes start out at two different airports. The planes are 2 044 km apart. The planes start flying towards each other. One plane is flying at 311 km/h and the other plane at 200 km/h.

If both planes started their journey at the same time, how long will they take to pass each other?

Answer: The two planes pass each other after hours.
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.


STEP: Sketch a simple picture to organise the information
[−2 points ⇒ 4 / 6 points left]

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two planes must be equal to the total distance when the planes meet:

d1+d2=dtotald1+d2=2 044 km

STEP: Write equations to describe the motion of the planes
[−2 points ⇒ 2 / 6 points left]

This question is about distances, speeds, and times. The equation connecting these values is

speed =distance time OR distance =speed ×time

We want to know the amount of time needed for the planes to meet. Let the time taken be t. Then we can write an expression for the distance each of the planes travels:

For plane 1:

d1=s1tplane is 311 km/hThe speed of the firstd1=311t

For plane 2:

d2=s2tplane is 200 km/hThe speed of the secondd2=200t

STEP: Solve the equations simultaneously for t
[−2 points ⇒ 0 / 6 points left]

Now we have three different equations. We can combine them using substitution to solve for the value of t.

d1+d2=2 044(311t)+(200t)=2 044511t=2 044t=2 044511t=4

The planes will meet after 4 hours.


Submit your answer as:

Word problems: distance, speed and time

Two trains start out at two different stations. The trains are 3 195 km apart. The trains start travelling towards each other. One train is travelling at 337 km/h and the other train at 302 km/h.

If both trains started their journey at the same time, how long will they take to pass each other?

Answer: The two trains pass each other after hours.
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.


STEP: Sketch a simple picture to organise the information
[−2 points ⇒ 4 / 6 points left]

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two trains must be equal to the total distance when the trains meet:

d1+d2=dtotald1+d2=3 195 km

STEP: Write equations to describe the motion of the trains
[−2 points ⇒ 2 / 6 points left]

This question is about distances, speeds, and times. The equation connecting these values is

speed =distance time OR distance =speed ×time

We want to know the amount of time needed for the trains to meet. Let the time taken be t. Then we can write an expression for the distance each of the trains travels:

For train 1:

d1=s1ttrain is 337 km/hThe speed of the firstd1=337t

For train 2:

d2=s2ttrain is 302 km/hThe speed of the secondd2=302t

STEP: Solve the equations simultaneously for t
[−2 points ⇒ 0 / 6 points left]

Now we have three different equations. We can combine them using substitution to solve for the value of t.

d1+d2=3 195(337t)+(302t)=3 195639t=3 195t=3 195639t=5

The trains will meet after 5 hours.


Submit your answer as:

Word problems: solving simultaneous equations

A group of friends is buying lunch together. The group buys 4 waffles and 5 sandwiches. Here are some facts about their lunch:

  • the total cost for the 4 waffles and 5 sandwiches is R276
  • a waffle costs R6 more than a sandwich

What is the price for one waffle and the price for one sandwich?

Answer:

A waffle costs R and a sandwich costs R .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

You need to choose variables to represent the things you want to find and write equations based on the information in the question.


STEP: Pick variables for the things we want to know
[−1 point ⇒ 6 / 7 points left]

The two things we want to know in this question are the prices for each of the items (a waffle and a sandwich). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

w=the price of a waffles=the price of a sandwich

STEP: Write equations based on the information in the question
[−2 points ⇒ 4 / 7 points left]

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the 4 waffles and 5 sandwiches is R276." We can use the expression 4w to represent the price of the 4 waffles. Similarly, the expression 5s represents the price of the 5 sandwiches. With these values we can write a full equation for the prices:

In words: the total cost for the 4 waffles and 5 sandwiches is R276In maths: 4w+5s=276

Now we can use the second point. It says, "a waffle costs R6 more than a sandwich." We can write this as an equation like this:

In words: a waffle costs R6 more than a sandwichIn maths: w=s+6

STEP: Solve the equations simultaneously
[−2 points ⇒ 2 / 7 points left]

Now that we have two equations, we need to solve them simultaneously.

4w+5s=276w=s+6

We could use elimination, but substitution is a better choice (because w is already the subject). Substitute the second equation into the first equation and solve!

4w+5s=2764(s+6)+5s=2764s+24+5s=2769s=27624s=2529=28

This means that the price of one sandwich is R28.


STEP: Find the other variable's value
[−1 point ⇒ 1 / 7 points left]

Finally, use the value we found for s to find the value of w (the price of a waffle). We can use either equation to do this, but the second one is easier to use (because w is already isolated).

w=s+6=28+6=34

The price for one waffle is R34.


STEP: Write the final answer
[−1 point ⇒ 0 / 7 points left]

It is important to write the answer to a word problem as a complete sentence.

The price of the waffle is R34 while a sandwich costs R28.


Submit your answer as: and

Word problems: solving simultaneous equations

A group of friends is buying lunch together. The group buys 4 sandwiches and 2 pizzas. Here are some facts about their lunch:

  • the total cost for the 4 sandwiches and 2 pizzas is R198
  • a sandwich costs R6 more than a pizza

What is the price for one sandwich and the price for one pizza?

Answer:

A sandwich costs R and a pizza costs R .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

You need to choose variables to represent the things you want to find and write equations based on the information in the question.


STEP: Pick variables for the things we want to know
[−1 point ⇒ 6 / 7 points left]

The two things we want to know in this question are the prices for each of the items (a sandwich and a pizza). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

s=the price of a sandwichp=the price of a pizza

STEP: Write equations based on the information in the question
[−2 points ⇒ 4 / 7 points left]

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the 4 sandwiches and 2 pizzas is R198." We can use the expression 4s to represent the price of the 4 sandwiches. Similarly, the expression 2p represents the price of the 2 pizzas. With these values we can write a full equation for the prices:

In words: the total cost for the 4 sandwiches and 2 pizzas is R198In maths: 4s+2p=198

Now we can use the second point. It says, "a sandwich costs R6 more than a pizza." We can write this as an equation like this:

In words: a sandwich costs R6 more than a pizzaIn maths: s=p+6

STEP: Solve the equations simultaneously
[−2 points ⇒ 2 / 7 points left]

Now that we have two equations, we need to solve them simultaneously.

4s+2p=198s=p+6

We could use elimination, but substitution is a better choice (because s is already the subject). Substitute the second equation into the first equation and solve!

4s+2p=1984(p+6)+2p=1984p+24+2p=1986p=19824p=1746=29

This means that the price of one pizza is R29.


STEP: Find the other variable's value
[−1 point ⇒ 1 / 7 points left]

Finally, use the value we found for p to find the value of s (the price of a sandwich). We can use either equation to do this, but the second one is easier to use (because s is already isolated).

s=p+6=29+6=35

The price for one sandwich is R35.


STEP: Write the final answer
[−1 point ⇒ 0 / 7 points left]

It is important to write the answer to a word problem as a complete sentence.

The price of the sandwich is R35 while a pizza costs R29.


Submit your answer as: and

Word problems: solving simultaneous equations

A group of friends is buying lunch together. The group buys 6 milkshakes and 4 waffles. Here are some facts about their lunch:

  • the total cost for the 6 milkshakes and 4 waffles is R264
  • a milkshake costs R4 more than a waffle

What is the price for one milkshake and the price for one waffle?

Answer:

A milkshake costs R and a waffle costs R .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

You need to choose variables to represent the things you want to find and write equations based on the information in the question.


STEP: Pick variables for the things we want to know
[−1 point ⇒ 6 / 7 points left]

The two things we want to know in this question are the prices for each of the items (a milkshake and a waffle). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

m=the price of a milkshakew=the price of a waffle

STEP: Write equations based on the information in the question
[−2 points ⇒ 4 / 7 points left]

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the 6 milkshakes and 4 waffles is R264." We can use the expression 6m to represent the price of the 6 milkshakes. Similarly, the expression 4w represents the price of the 4 waffles. With these values we can write a full equation for the prices:

In words: the total cost for the 6 milkshakes and 4 waffles is R264In maths: 6m+4w=264

Now we can use the second point. It says, "a milkshake costs R4 more than a waffle." We can write this as an equation like this:

In words: a milkshake costs R4 more than a waffleIn maths: m=w+4

STEP: Solve the equations simultaneously
[−2 points ⇒ 2 / 7 points left]

Now that we have two equations, we need to solve them simultaneously.

6m+4w=264m=w+4

We could use elimination, but substitution is a better choice (because m is already the subject). Substitute the second equation into the first equation and solve!

6m+4w=2646(w+4)+4w=2646w+24+4w=26410w=26424w=24010=24

This means that the price of one waffle is R24.


STEP: Find the other variable's value
[−1 point ⇒ 1 / 7 points left]

Finally, use the value we found for w to find the value of m (the price of a milkshake). We can use either equation to do this, but the second one is easier to use (because m is already isolated).

m=w+4=24+4=28

The price for one milkshake is R28.


STEP: Write the final answer
[−1 point ⇒ 0 / 7 points left]

It is important to write the answer to a word problem as a complete sentence.

The price of the milkshake is R28 while a waffle costs R24.


Submit your answer as: and

Word problems: checking answers about shopping

Suppose you must solve this word problem:

At a shop there are some fruits and sweets for sale. Ayodeji buys some bananas and some chocolates. The total cost for 5 bananas and 5 chocolates is R48,00. And each chocolate costs R0,60 less than each banana. What is the price for one banana?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the price be a decimal number (or must it be an integer)?
  2. Can the price be a negative number?
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the price of each banana. Can the price be a decimal, like 4,5? Can it be negative, like 5?


STEP: Decide if the price can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be R6,75. Or R5. Or R19,87. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the price can be a decimal number or not. In other words, can the price of something be a decimal number like R4,30, or must a price always be a whole number like R4,00? Of course we know that many prices are not whole numbers! This is similar to the example above about airtime, which can be a decimal number. There is no reason why the price of an item must be a whole number.

The answer for the first question is: Yes, the price can be a decimal number.


STEP: Decide if the price can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if the price can be negative, or must be positive. Prices cannot be negative. Imagine something which costs -R5,50. That is impossible!

The answer to the second question is: No, the price cannot be a negative number.

The correct answer choices are:

  1. Yes, it can be a decimal
  2. No, it cannot be negative

Submit your answer as: and

Word problems: checking answers about shopping

Suppose you must solve this word problem:

At a shop there are some fruits and sweets for sale. Yaseen buys some mangoes and some packs of gum. The total cost for 2 mangoes and 4 packs of gum is R34,80. And each pack of gum costs R0,30 less than each mango. What is the price for one mango?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the price be a decimal number (or must it be an integer)?
  2. Can the price be a negative number?
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the price of each mango. Can the price be a decimal, like 4,5? Can it be negative, like 5?


STEP: Decide if the price can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be R6,75. Or R5. Or R19,87. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the price can be a decimal number or not. In other words, can the price of something be a decimal number like R4,30, or must a price always be a whole number like R4,00? Of course we know that many prices are not whole numbers! This is similar to the example above about airtime, which can be a decimal number. There is no reason why the price of an item must be a whole number.

The answer for the first question is: Yes, the price can be a decimal number.


STEP: Decide if the price can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if the price can be negative, or must be positive. Prices cannot be negative. Imagine something which costs -R5,50. That is impossible!

The answer to the second question is: No, the price cannot be a negative number.

The correct answer choices are:

  1. Yes, it can be a decimal
  2. No, it cannot be negative

Submit your answer as: and

Word problems: checking answers about shopping

Suppose you must solve this word problem:

At a shop there are some fruits and sweets for sale. Ismail buys some bananas and some packs of gum. The total cost for 2 bananas and 5 packs of gum is R29,10. And each pack of gum costs R2,30 less than each banana. What is the price for one banana?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the price be a decimal number (or must it be an integer)?
  2. Can the price be a negative number?
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the price of each banana. Can the price be a decimal, like 4,5? Can it be negative, like 5?


STEP: Decide if the price can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be R6,75. Or R5. Or R19,87. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the price can be a decimal number or not. In other words, can the price of something be a decimal number like R4,30, or must a price always be a whole number like R4,00? Of course we know that many prices are not whole numbers! This is similar to the example above about airtime, which can be a decimal number. There is no reason why the price of an item must be a whole number.

The answer for the first question is: Yes, the price can be a decimal number.


STEP: Decide if the price can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if the price can be negative, or must be positive. Prices cannot be negative. Imagine something which costs -R5,50. That is impossible!

The answer to the second question is: No, the price cannot be a negative number.

The correct answer choices are:

  1. Yes, it can be a decimal
  2. No, it cannot be negative

Submit your answer as: and

Word problems: checking answers with shapes

Suppose you must solve this word problem:

Ntombebhongo is drawing a pattern of rectangles in her notebook. The number of rectangles in each figure is 3 more than the number of rectangles in the previous figure. If there are 12 rectangles in the third figure, how many rectangles are there in the first figure?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the number of rectangles be a decimal number (or must it be an integer)?
  2. Can the number of rectangles be a negative number?
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the number of rectangles in the first figure of Ntombebhongo's diagram. Can the number of rectangles be a decimal, like 4,5? Can it be negative, like 5?


STEP: Decide if the number of rectangles can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be 150 days of school, shame. Similarly, it is not possible to have 176,3562 days of school. What type of numbers are realistic for the number of rectangles?

For the first question we must decide if the number of rectangles can be a decimal number. Is that fine or is it impossible?

In other words, can the number of rectangles be a decimal number like 3,7, or must the number be a whole number like 4? Just like the number of days in the school year, the number of rectangles cannot be a decimal value. 3,7 rectangles does not make sense! The number of rectangles must be a whole number! (You might think, "That's wrong, we can just draw part of a rectangle to get 3,7 rectangles." But if the shape is not complete, it is not a rectangle.)

The answer for the first question is: No, it must be an integer, the number of rectangles cannot be a decimal number.


STEP: Decide if the number of rectangles can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if number of rectangles can be negative, or must be positive. The number of rectangles cannot be negative. If Ntombebhongo drew any rectangles at all, then the answer to this question must be positive. There is no such thing as 4 rectangles.

The answer for the second question is: No, the number of rectangles cannot be a negative number.

The correct answer choices are:

  1. No, it must be an integer
  2. No, it cannot be negative

Submit your answer as: and

Word problems: checking answers with shapes

Suppose you must solve this word problem:

Chris is drawing a rectangle in his notebook. The length of the rectangle is 5 cm more than its width. The area of the rectangle is 6 cm2. What is the width of the rectangle?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the width of the rectangle be a decimal number (or must it be an integer)?
  2. Can the width of the rectangle be a negative number?
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the width of the rectangle. Can the width of the rectangle be a decimal, like 4,5? Can it be negative, like 5?


STEP: Decide if the width of the rectangle can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be 150 days of school, shame. Similarly, it is not possible to have 176,3562 days of school. What type of numbers are realistic for the width of the rectangle?

For the first question we must decide if the width of the rectangle can be a decimal number. Is that fine or is it impossible?

In other words, can a distance be a decimal number like 3,5 cm, or must a distance be a whole number like 4 cm? The answer is that there is no reason a distance must be limited to whole number values. This is different from the number of days in the school year. It is not possible to have 3,5 days of school, but a rectangle certainly can have a width of 3,5 cm. Or 5,927 cm. There is no reason why the width of the rectangle must be a whole number.

The answer for the first question is: Yes, it can be a decimal, the width of the rectangle can be a decimal number.


STEP: Decide if the width of the rectangle can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if width of the rectangle can be negative, or must be positive. This is straightfoward: distances cannot be negative. Imagine a room 5 metres wide. No way!

The answer for the second question is: No, the width of the rectangle cannot be a negative number.

The correct answer choices are:

  1. Yes, it can be a decimal
  2. No, it cannot be negative

Submit your answer as: and

Word problems: checking answers with shapes

Suppose you must solve this word problem:

Habubakar is drawing a pattern of rectangles in his notebook. The number of rectangles in each figure is 2 more than the number of rectangles in the previous figure. If there are 11 rectangles in the fourth figure, how many rectangles are there in the first figure?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the number of rectangles be a decimal number (or must it be an integer)?
  2. Can the number of rectangles be a negative number?
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the number of rectangles in the first figure of Habubakar's diagram. Can the number of rectangles be a decimal, like 4,5? Can it be negative, like 5?


STEP: Decide if the number of rectangles can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be 150 days of school, shame. Similarly, it is not possible to have 176,3562 days of school. What type of numbers are realistic for the number of rectangles?

For the first question we must decide if the number of rectangles can be a decimal number. Is that fine or is it impossible?

In other words, can the number of rectangles be a decimal number like 3,7, or must the number be a whole number like 4? Just like the number of days in the school year, the number of rectangles cannot be a decimal value. 3,7 rectangles does not make sense! The number of rectangles must be a whole number! (You might think, "That's wrong, we can just draw part of a rectangle to get 3,7 rectangles." But if the shape is not complete, it is not a rectangle.)

The answer for the first question is: No, it must be an integer, the number of rectangles cannot be a decimal number.


STEP: Decide if the number of rectangles can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if number of rectangles can be negative, or must be positive. The number of rectangles cannot be negative. If Habubakar drew any rectangles at all, then the answer to this question must be positive. There is no such thing as 4 rectangles.

The answer for the second question is: No, the number of rectangles cannot be a negative number.

The correct answer choices are:

  1. No, it must be an integer
  2. No, it cannot be negative

Submit your answer as: and

Word problems: test scores and simultaneous equations

Louis and Musa are friends. Louis takes Musa's physics test paper and says: “I have 14 marks more than you do and the sum of both our marks is equal to 166. What are our marks?”

Answer:

Louis got marks and Musa got marks for the physics test paper.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.


STEP: Pick variables for the marks
[−1 point ⇒ 5 / 6 points left]

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

l= Louis's markm= Musa's mark

STEP: Write equations about the students' marks
[−2 points ⇒ 3 / 6 points left]

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

 more than MusaLouis has 14 marksl=m+14(the total) is 166The sum of the marksl+m=166

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

l+m=166(m+14)+m=1662m=16614m=1522=76

This means that Musa's mark is 76.


STEP: Find Louis's mark
[−1 point ⇒ 0 / 6 points left]

Now we can use Musa's mark and one of the equations we have to find Louis's mark. The easiest way to do that is to substitute Musa's mark back into the first equation:

l=m+14=(76)+14=90

The students achieved these marks: Louis earned 90 and Musa earned 76.


Submit your answer as: and

Word problems: test scores and simultaneous equations

Nicole and Gabisile are friends. Nicole takes Gabisile's chemistry test paper and says: “I have 19 marks more than you do and the sum of both our marks is equal to 117. What are our marks?”

Answer:

Nicole got marks and Gabisile got marks for the chemistry test paper.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.


STEP: Pick variables for the marks
[−1 point ⇒ 5 / 6 points left]

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

n= Nicole's markg= Gabisile's mark

STEP: Write equations about the students' marks
[−2 points ⇒ 3 / 6 points left]

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

 more than GabisileNicole has 19 marksn=g+19(the total) is 117The sum of the marksn+g=117

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

n+g=117(g+19)+g=1172g=11719g=982=49

This means that Gabisile's mark is 49.


STEP: Find Nicole's mark
[−1 point ⇒ 0 / 6 points left]

Now we can use Gabisile's mark and one of the equations we have to find Nicole's mark. The easiest way to do that is to substitute Gabisile's mark back into the first equation:

n=g+19=(49)+19=68

The students achieved these marks: Nicole earned 68 and Gabisile earned 49.


Submit your answer as: and

Word problems: test scores and simultaneous equations

Bianca and Carl are friends. Bianca takes Carl's mathematics test paper and says: “I have 19 marks more than you do and the sum of both our marks is equal to 153. What are our marks?”

Answer:

Bianca got marks and Carl got marks for the mathematics test paper.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.


STEP: Pick variables for the marks
[−1 point ⇒ 5 / 6 points left]

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

b= Bianca's markc= Carl's mark

STEP: Write equations about the students' marks
[−2 points ⇒ 3 / 6 points left]

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

 more than CarlBianca has 19 marksb=c+19(the total) is 153The sum of the marksb+c=153

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

b+c=153(c+19)+c=1532c=15319c=1342=67

This means that Carl's mark is 67.


STEP: Find Bianca's mark
[−1 point ⇒ 0 / 6 points left]

Now we can use Carl's mark and one of the equations we have to find Bianca's mark. The easiest way to do that is to substitute Carl's mark back into the first equation:

b=c+19=(67)+19=86

The students achieved these marks: Bianca earned 86 and Carl earned 67.


Submit your answer as: and

Word problems: odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive odd integers.
  • The sum of the numbers is 24.

What is the value of the smaller number?

Answer: The smaller number is .
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. For example, you can use n1 and n2. Then write equations using these variables based on the facts given in the question.
STEP: Pick variables
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find one of them (the smaller one). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a sum of 24. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

we need to findthis is the numbern1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact given in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, 11 and 13 are consecutive odd numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the even integer which sits between n1 and n2.

The second fact about the numbers tells us that "the sum of the numbers is 24". Remember that sum means addition. So:

n1+n2=24

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

We can do this using substitution. If we substitute n2=n1+2 into the equation n1+n2=24, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1+(n1+2)=242n1+2=242n1=22n1=11

The result is n1=11. Notice that this means that the other number, n2, must be 13, because n2=n1+2. This is perfect, because we know that the sum of the numbers is 24, and 11+13=24.

The smaller number is 11.


Submit your answer as:

Word problems: odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive even integers.
  • The sum of the numbers is 30.

Determine the value of the smaller number.

Answer: The smaller number is .
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. For example, you can use n1 and n2. Then write equations using these variables based on the facts given in the question.
STEP: Pick variables
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find one of them (the smaller one). We know certain things about these numbers: they are positive, they are consecutive even integers, and they have a sum of 30. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

we need to findthis is the numbern1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact given in the question says that the numbers are "consecutive even integers". So both of the numbers are even, and they come one after another. For example, 10 and 12 are consecutive even numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the odd integer which sits between n1 and n2.

The second fact about the numbers tells us that "the sum of the numbers is 30". Remember that sum means addition. So:

n1+n2=30

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

We can do this using substitution. If we substitute n2=n1+2 into the equation n1+n2=30, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1+(n1+2)=302n1+2=302n1=28n1=14

The result is n1=14. Notice that this means that the other number, n2, must be 16, because n2=n1+2. This is perfect, because we know that the sum of the numbers is 30, and 14+16=30.

The smaller number is 14.


Submit your answer as:

Word problems: odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive odd integers.
  • The sum of the numbers is 32.

What is the value of the larger number?

Answer: The larger number is .
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. For example, you can use n1 and n2. Then write equations using these variables based on the facts given in the question.
STEP: Pick variables
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find one of them (the larger one). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a sum of 32. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

n1=the smaller numberwe need to findthis is the numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact given in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, 11 and 13 are consecutive odd numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the even integer which sits between n1 and n2.

The second fact about the numbers tells us that "the sum of the numbers is 32". Remember that sum means addition. So:

n1+n2=32

STEP: Solve the equations for n2
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the larger number (we do not need both of them). That means we need to find the value of n2.

We can solve this using substitution. However, remember that we want the value of n2 (the larger number). We can start by rearranging the first equation to make n1 the subject. Then the substitution step will remove n1 from the second equation and we can solve for n2. (This is not required - it just makes the solution faster.)

n2=n1+2n22=n1

Now substitute n22 into the other equation and solve for n2.

(n22)+n2=322n22=322n2=34n2=17

The result is n2=17. Notice that this means that the other number, n1, must be 15, because n2=n1+2. This is perfect, because we know that the sum of the numbers is 32, and 17+15=32.

The larger number is 17.


Submit your answer as:

Word problems: checking answers

Balarabe is 9 years younger than her sister, Nyakallo. In 5 years, Nyakallo will be 4 times as old as Balarabe. How old is Balarabe now?

INSTRUCTION: If there is no acceptable solution, type 'no solution' in the answer box.
Answer:

Balarabe is years old.

one-of
type(string.nocase)
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to find Balarabe's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.


STEP: Choose variables for each person's age
[−1 point ⇒ 4 / 5 points left]

We need to determine Balarabe's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Nyakallo and Balarabe. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

aN=Nyakallo's ageaB=Balarabe's age

STEP: Write an equation based on the information in the question
[−1 point ⇒ 3 / 5 points left]

Now let's write equations using these variables. From the question we know that: "Balarabe is 9 years younger than her sister, Nyakallo." We need to translate that into mathematics. The key word is younger, which tells us to use subtraction to relate the ages.

aB=aN9

STEP: Write the equations
[−1 point ⇒ 2 / 5 points left]

We also know that: "In 5 years, Nyakallo will be 4 times as old as Balarabe." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

aN+5=Nyakallo's age in 5 yearsaB+5=Balarabe's age in 5 years

These are the ages at which Nyakallo will be 4 times as old as Balarabe. We can put all this together as follows:

in 5 yearsNyakallo's age=4×in 5 yearsBalarabe's ageaN+5=4(aB+5)

STEP: Solve the equations simultaneously
[−2 points ⇒ 0 / 5 points left]

Now we have to solve two equations, both of them including the variables aN and aB. The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is aB=aN9. We can substitute this into the second equation and solve for aN.

aN+5=4(aB+5)aN+5=4((aN9)+5)aN+5=4aN1621=3aN7=aN

Terrific: this means that Nyakallo is 7 years old. But the question asked for us to find Balarabe's age. We can find it using the first equation, which relates the two ages:

aB=aN9=(7)9=2

So we finally got the answer to the question. But wait a minute: aB represents the age of a person. It can't be negative! This either means that we made a mistake, or that there is no solution to the question. There is no mistake in the work: it turns out that the facts given about the two people's ages are not possible! The numbers agree with all the relationships given in the question, but we cannot forget that the numbers in this question have meaning. They refer to how many years it has been since someone was born. And that cannot be a negative number.

The correct answer is: no solution.


Submit your answer as:

Word problems: checking answers

Chinweike is 12 years younger than her sister, Joanna. In 8 years, Joanna will be 2 times as old as Chinweike. How old is Chinweike now?

INSTRUCTION: If there is no acceptable solution, type 'no solution' in the answer box.
Answer:

Chinweike is years old.

numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to find Chinweike's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.


STEP: Choose variables for each person's age
[−1 point ⇒ 4 / 5 points left]

We need to determine Chinweike's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Joanna and Chinweike. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

aJ=Joanna's ageaC=Chinweike's age

STEP: Write an equation based on the information in the question
[−1 point ⇒ 3 / 5 points left]

Now let's write equations using these variables. From the question we know that: "Chinweike is 12 years younger than her sister, Joanna." We need to translate that into mathematics. The key word is younger, which tells us to use subtraction to relate the ages.

aC=aJ12

STEP: Write the equations
[−1 point ⇒ 2 / 5 points left]

We also know that: "In 8 years, Joanna will be 2 times as old as Chinweike." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

aJ+8=Joanna's age in 8 yearsaC+8=Chinweike's age in 8 years

These are the ages at which Joanna will be 2 times as old as Chinweike. We can put all this together as follows:

in 8 yearsJoanna's age=2×in 8 yearsChinweike's ageaJ+8=2(aC+8)

STEP: Solve the equations simultaneously
[−2 points ⇒ 0 / 5 points left]

Now we have to solve two equations, both of them including the variables aJ and aC. The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is aC=aJ12. We can substitute this into the second equation and solve for aJ.

aJ+8=2(aC+8)aJ+8=2((aJ12)+8)aJ+8=2aJ816=aJ

Terrific: this means that Joanna is 16 years old. But the question asked for us to find Chinweike's age. We can find it using the first equation, which relates the two ages:

aC=aJ12=(16)12=4

So we finally got the answer to the question. Chinweike is 4 years old.

The correct answer is: 4.


Submit your answer as:

Word problems: checking answers

Deborah is 11 years younger than her sister, Qiniso. In 9 years, Qiniso will be 12 times as old as Deborah. How old is Deborah now?

INSTRUCTION: If there is no acceptable solution, type 'no solution' in the answer box.
Answer:

Deborah is years old.

one-of
type(string.nocase)
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to find Deborah's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.


STEP: Choose variables for each person's age
[−1 point ⇒ 4 / 5 points left]

We need to determine Deborah's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Qiniso and Deborah. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

aQ=Qiniso's ageaD=Deborah's age

STEP: Write an equation based on the information in the question
[−1 point ⇒ 3 / 5 points left]

Now let's write equations using these variables. From the question we know that: "Deborah is 11 years younger than her sister, Qiniso." We need to translate that into mathematics. The key word is younger, which tells us to use subtraction to relate the ages.

aD=aQ11

STEP: Write the equations
[−1 point ⇒ 2 / 5 points left]

We also know that: "In 9 years, Qiniso will be 12 times as old as Deborah." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

aQ+9=Qiniso's age in 9 yearsaD+9=Deborah's age in 9 years

These are the ages at which Qiniso will be 12 times as old as Deborah. We can put all this together as follows:

in 9 yearsQiniso's age=12×in 9 yearsDeborah's ageaQ+9=12(aD+9)

STEP: Solve the equations simultaneously
[−2 points ⇒ 0 / 5 points left]

Now we have to solve two equations, both of them including the variables aQ and aD. The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is aD=aQ11. We can substitute this into the second equation and solve for aQ.

aQ+9=12(aD+9)aQ+9=12((aQ11)+9)aQ+9=12aQ2433=11aQ3=aQ

Terrific: this means that Qiniso is 3 years old. But the question asked for us to find Deborah's age. We can find it using the first equation, which relates the two ages:

aD=aQ11=(3)11=8

So we finally got the answer to the question. But wait a minute: aD represents the age of a person. It can't be negative! This either means that we made a mistake, or that there is no solution to the question. There is no mistake in the work: it turns out that the facts given about the two people's ages are not possible! The numbers agree with all the relationships given in the question, but we cannot forget that the numbers in this question have meaning. They refer to how many years it has been since someone was born. And that cannot be a negative number.

The correct answer is: no solution.


Submit your answer as:

Word problems: products of odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive even integers.
  • The product of the numbers is 80.

Find the value of the larger number.

Answer: The larger number is .
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. Then write equations using these variables based on the information given in the question.
STEP: Pick variables for the numbers
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find the larger number (not both). We know certain things about these numbers: they are positive, they are consecutive even integers, and they have a product of 80. We can use that information to solve this question using simultaneous equations.

The first thing we need to do is define variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent the things we want to find. In this case we are looking for two numbers, so these are good choices:

n1=the smaller numberwe need to findthis is the numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact in the question says that the numbers are "consecutive even integers". So both of the numbers are even, and they come one after another. For example, 10 and 12 are consecutive even numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the odd integer which sits between n1 and n2.

The second fact about the numbers tells us that "the product of the numbers is 80". Remember that product means multiplication. So:

n1n2=80

STEP: Solve the equations for n2
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the larger number (we do not need both of them). That means we need to find the value of n2.

Remember that we want the value of n2 (the larger number). We can start by rearranging the first equation to make n1 the subject. Then the substitution step will remove n1 from the second equation and we can solve for n2. (This is not required - it just makes the solution faster.)

n2=n1+2n22=n1

Now substitute n22 into the other equation and solve for n2.

(n22)n2=80n222n2=80n222n280=0(n210)(n2+8)=0
n2=10 and n2=8

This solution led to two answers for n2. However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves n2=10.

This means the other number, n1, must be 8, because n2=n1+2. This is perfect, because we also know that the product of the numbers is 80, and 108=80.

The larger number is 10.


Submit your answer as:

Word problems: products of odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive odd integers.
  • The product of the numbers is 63.

Find the value of the larger number.

Answer: The larger number is .
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. Then write equations using these variables based on the information given in the question.
STEP: Pick variables for the numbers
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find the larger number (not both). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a product of 63. We can use that information to solve this question using simultaneous equations.

The first thing we need to do is define variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent the things we want to find. In this case we are looking for two numbers, so these are good choices:

n1=the smaller numberwe need to findthis is the numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, 11 and 13 are consecutive odd numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the even integer which sits between n1 and n2.

The second fact about the numbers tells us that "the product of the numbers is 63". Remember that product means multiplication. So:

n1n2=63

STEP: Solve the equations for n2
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the larger number (we do not need both of them). That means we need to find the value of n2.

Remember that we want the value of n2 (the larger number). We can start by rearranging the first equation to make n1 the subject. Then the substitution step will remove n1 from the second equation and we can solve for n2. (This is not required - it just makes the solution faster.)

n2=n1+2n22=n1

Now substitute n22 into the other equation and solve for n2.

(n22)n2=63n222n2=63n222n263=0(n29)(n2+7)=0
n2=9 and n2=7

This solution led to two answers for n2. However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves n2=9.

This means the other number, n1, must be 7, because n2=n1+2. This is perfect, because we also know that the product of the numbers is 63, and 97=63.

The larger number is 9.


Submit your answer as:

Word problems: products of odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive odd integers.
  • The product of the numbers is 3.

What is the value of the smaller number?

Answer: The smaller number is .
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. Then write equations using these variables based on the information given in the question.
STEP: Pick variables for the numbers
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find the smaller number (not both). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a product of 3. We can use that information to solve this question using simultaneous equations.

The first thing we need to do is define variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent the things we want to find. In this case we are looking for two numbers, so these are good choices:

we need to findthis is the numbern1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, 11 and 13 are consecutive odd numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the even integer which sits between n1 and n2.

The second fact about the numbers tells us that "the product of the numbers is 3". Remember that product means multiplication. So:

n1n2=3

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

If we subsitute n2=n1+2 into the equation n1n2=3, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1(n1+2)=3n12+2n1=3n12+2n13=0(n1+3)(n11)=0
n1=3andn1=1

This solution led to two answers for n1. However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves n1=1.

This means the other number, n2, must be 3, because n2=n1+2. This is perfect, because we also know that the product of the numbers is 3, and 13=3.

The smaller number is 1.


Submit your answer as:

Word problems: finding consecutive numbers

Here are some facts about two numbers:

  • the numbers are consecutive integers
  • both numbers are negative
  • the product of the numbers is 30

What are the two numbers?

TIP: 'Product' means multiplication; 'consecutive' means 'following each other without gaps between.'
INSTRUCTION: Write your answers in the boxes below. It does not matter what order your answers are in.
Answer: The numbers are and .
numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You should represent the numbers in the question by n and n+1. Write an equation based on the fact that the product of those two numbers is 30.


STEP: Pick variables to represent the numbers
[−1 point ⇒ 5 / 6 points left]

This question asks us to find two numbers which have a product of 30. That means if we multiply the numbers we get 30.

The first thing we need to do is pick variables for the numbers we are trying to find. We can pick any variable. But it is a good idea to pick a variable which is related to whatever it represents. In this question, n is a good choice because we are looking for numbers. In fact, since there are two different numbers, we can do this:

n1=the first numbern2=the next number

STEP: Write an equation based on the information in the question
[−1 point ⇒ 4 / 6 points left]

The question tells us that the product of these numbers is 30. In other words, if we multiply the numbers, we get 30. As an equation this is:

n1n2=30

STEP: Connect the two variables
[−1 point ⇒ 3 / 6 points left]

Now we can use the fact that the numbers are consecutive. So if n1 represents the first number, then the next number must be one more than n1.

If:n2=the number following n1then:n2=n1+1

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have two equations about the two numbers.

n1n2=30n2=n1+1

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for n1.

n1(n1+1)=30(n1)2+n130=0(n1+6)(n15)=0
n1=6andn1=5

The solutions to the equation are n1=6 or n1=5. This means the first number is either −6 or 5.


STEP: Find the final answers
[−1 point ⇒ 0 / 6 points left]

Wait a minute! The numbers −6 and 5 are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are negative. That means we must throw out the n1=5, so the first number is n1=6.

Based on that we can find the second number. The second number is n1+1, which is equal to (6)+1=5.

The two consecutive integers are −6 and −5.


Submit your answer as: and

Word problems: finding consecutive numbers

Here are some facts about two numbers:

  • the numbers are consecutive integers
  • both numbers are positive
  • the product of the numbers is 110

What are the two numbers?

TIP: 'Product' means multiplication; 'consecutive' means 'following each other without gaps between.'
INSTRUCTION: Write your answers in the boxes below. It does not matter what order your answers are in.
Answer: The numbers are and .
numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You should represent the numbers in the question by n and n+1. Write an equation based on the fact that the product of those two numbers is 110.


STEP: Pick variables to represent the numbers
[−1 point ⇒ 5 / 6 points left]

This question asks us to find two numbers which have a product of 110. That means if we multiply the numbers we get 110.

The first thing we need to do is pick variables for the numbers we are trying to find. We can pick any variable. But it is a good idea to pick a variable which is related to whatever it represents. In this question, n is a good choice because we are looking for numbers. In fact, since there are two different numbers, we can do this:

n1=the first numbern2=the next number

STEP: Write an equation based on the information in the question
[−1 point ⇒ 4 / 6 points left]

The question tells us that the product of these numbers is 110. In other words, if we multiply the numbers, we get 110. As an equation this is:

n1n2=110

STEP: Connect the two variables
[−1 point ⇒ 3 / 6 points left]

Now we can use the fact that the numbers are consecutive. So if n1 represents the first number, then the next number must be one more than n1.

If:n2=the number following n1then:n2=n1+1

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have two equations about the two numbers.

n1n2=110n2=n1+1

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for n1.

n1(n1+1)=110(n1)2+n1110=0(n1+11)(n110)=0
n1=11andn1=10

The solutions to the equation are n1=11 or n1=10. This means the first number is either −11 or 10.


STEP: Find the final answers
[−1 point ⇒ 0 / 6 points left]

Wait a minute! The numbers −11 and 10 are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are positive. That means we must throw out the n1=11, so the first number is n1=10.

Based on that we can find the second number. The second number is n1+1, which is equal to (10)+1=11.

The two consecutive integers are 10 and 11.


Submit your answer as: and

Word problems: finding consecutive numbers

Here are some facts about two numbers:

  • the numbers are consecutive integers
  • both numbers are negative
  • the product of the numbers is 306

What are the two numbers?

TIP: 'Product' means multiplication; 'consecutive' means 'following each other without gaps between.'
INSTRUCTION: Write your answers in the boxes below. It does not matter what order your answers are in.
Answer: The numbers are and .
numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You should represent the numbers in the question by n and n+1. Write an equation based on the fact that the product of those two numbers is 306.


STEP: Pick variables to represent the numbers
[−1 point ⇒ 5 / 6 points left]

This question asks us to find two numbers which have a product of 306. That means if we multiply the numbers we get 306.

The first thing we need to do is pick variables for the numbers we are trying to find. We can pick any variable. But it is a good idea to pick a variable which is related to whatever it represents. In this question, n is a good choice because we are looking for numbers. In fact, since there are two different numbers, we can do this:

n1=the first numbern2=the next number

STEP: Write an equation based on the information in the question
[−1 point ⇒ 4 / 6 points left]

The question tells us that the product of these numbers is 306. In other words, if we multiply the numbers, we get 306. As an equation this is:

n1n2=306

STEP: Connect the two variables
[−1 point ⇒ 3 / 6 points left]

Now we can use the fact that the numbers are consecutive. So if n1 represents the first number, then the next number must be one more than n1.

If:n2=the number following n1then:n2=n1+1

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have two equations about the two numbers.

n1n2=306n2=n1+1

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for n1.

n1(n1+1)=306(n1)2+n1306=0(n1+18)(n117)=0
n1=18andn1=17

The solutions to the equation are n1=18 or n1=17. This means the first number is either −18 or 17.


STEP: Find the final answers
[−1 point ⇒ 0 / 6 points left]

Wait a minute! The numbers −18 and 17 are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are negative. That means we must throw out the n1=17, so the first number is n1=18.

Based on that we can find the second number. The second number is n1+1, which is equal to (18)+1=17.

The two consecutive integers are −18 and −17.


Submit your answer as: and

Word problems: an age-old question

Jacob has a son, Chinedu. Here are some facts about how old Jacob and Chinedu are:

  • Jacob is 9 times as old as Chinedu right now.
  • 5 years from now, Jacob will be 4 times as old as Chinedu.

How old are Jacob and Chinedu now?

Answer:

Jacob is years old and Chinedu is years old.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.


STEP: Choose variables for the information
[−1 point ⇒ 5 / 6 points left]

In this question we want to find the age of two people: Jacob and his son, Chinedu. We can solve this by setting up two equations and solving them simultaneously.

Start by choosing variables to represent the ages of the father and the son. (We need to do this because we don't know the ages!) It is a good idea to choose variables that match what we are describing. So:

Let j=Jacob's ageLet c=Chinedu's age

STEP: Write equations based on the information in the question
[−2 points ⇒ 3 / 6 points left]

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Jacob is now 9 times as old as Chinedu." As an equation, this is:

Ages now: j=9c

The second piece of information says that in "5 years... Jacob will be 4 times as old as his son." In 5 years Jacob will be j+5 years old, and similarly Chinedu will be c+5 years old. Then:

Ages in 5 years: j+5=4(c+5)

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have a pair of simultaneous equations! Substitute the first equation into the second equation and solve. (You can solve the equations using elimination if you prefer. In that case, the easiest option is to subtract the equations to cancel j.)

j+5=4(c+5)(9c)+5=4c+205c=15c=3

Great! Now we know that Chinedu is 3 years old.


STEP: Use Chinedu's age to find Jacob's age
[−1 point ⇒ 0 / 6 points left]

We can now find Jacob's age. Substitute Chinedu's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

j=9c=9(3)=27

Write your final answer: Jacob is 27 years old and Chinedu is 3 years old.


Submit your answer as: and

Word problems: an age-old question

Ben has a son, Gift. Here are some facts about how old Ben and Gift are:

  • Ben is 11 times as old as Gift right now.
  • 4 years from now, Ben will be 6 times as old as Gift.

How old are Ben and Gift now?

Answer:

Ben is years old and Gift is years old.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.


STEP: Choose variables for the information
[−1 point ⇒ 5 / 6 points left]

In this question we want to find the age of two people: Ben and his son, Gift. We can solve this by setting up two equations and solving them simultaneously.

Start by choosing variables to represent the ages of the father and the son. (We need to do this because we don't know the ages!) It is a good idea to choose variables that match what we are describing. So:

Let b=Ben's ageLet g=Gift's age

STEP: Write equations based on the information in the question
[−2 points ⇒ 3 / 6 points left]

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Ben is now 11 times as old as Gift." As an equation, this is:

Ages now: b=11g

The second piece of information says that in "4 years... Ben will be 6 times as old as his son." In 4 years Ben will be b+4 years old, and similarly Gift will be g+4 years old. Then:

Ages in 4 years: b+4=6(g+4)

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have a pair of simultaneous equations! Substitute the first equation into the second equation and solve. (You can solve the equations using elimination if you prefer. In that case, the easiest option is to subtract the equations to cancel b.)

b+4=6(g+4)(11g)+4=6g+245g=20g=4

Great! Now we know that Gift is 4 years old.


STEP: Use Gift's age to find Ben's age
[−1 point ⇒ 0 / 6 points left]

We can now find Ben's age. Substitute Gift's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

b=11g=11(4)=44

Write your final answer: Ben is 44 years old and Gift is 4 years old.


Submit your answer as: and

Word problems: an age-old question

Sehlolo has a son, Peter. Here are some facts about how old Sehlolo and Peter are:

  • Sehlolo is 9 times as old as Peter right now.
  • 9 years from now, Sehlolo will be 3 times as old as Peter.

How old are Sehlolo and Peter now?

Answer:

Sehlolo is years old and Peter is years old.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.


STEP: Choose variables for the information
[−1 point ⇒ 5 / 6 points left]

In this question we want to find the age of two people: Sehlolo and his son, Peter. We can solve this by setting up two equations and solving them simultaneously.

Start by choosing variables to represent the ages of the father and the son. (We need to do this because we don't know the ages!) It is a good idea to choose variables that match what we are describing. So:

Let s=Sehlolo's ageLet p=Peter's age

STEP: Write equations based on the information in the question
[−2 points ⇒ 3 / 6 points left]

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Sehlolo is now 9 times as old as Peter." As an equation, this is:

Ages now: s=9p

The second piece of information says that in "9 years... Sehlolo will be 3 times as old as his son." In 9 years Sehlolo will be s+9 years old, and similarly Peter will be p+9 years old. Then:

Ages in 9 years: s+9=3(p+9)

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have a pair of simultaneous equations! Substitute the first equation into the second equation and solve. (You can solve the equations using elimination if you prefer. In that case, the easiest option is to subtract the equations to cancel s.)

s+9=3(p+9)(9p)+9=3p+276p=18p=3

Great! Now we know that Peter is 3 years old.


STEP: Use Peter's age to find Sehlolo's age
[−1 point ⇒ 0 / 6 points left]

We can now find Sehlolo's age. Substitute Peter's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

s=9p=9(3)=27

Write your final answer: Sehlolo is 27 years old and Peter is 3 years old.


Submit your answer as: and

Word Problems: checking answers

Suppose you must solve this word problem:

A marathon is a long-distance running race with an official distance of 42,195 kilometres. 1 000 people ran the "It Ain't Over Till It's Over" marathon in Port Elizabeth. The second-place runner finished 19,82 seconds after the winner. The third runner was 16,64 seconds behind the second runner, and the fourth runner was 24,21 seconds behind the third runner. If the second-place runner had an average speed of 18,72 km/h, what was the winner's average speed for the marathon?

Can the answer to the question above be a negative number?

Answer:

Can the answer be negative?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Think about the meaning of the number you need. Some values cannot be decimal numbers. Other values cannot be negative numbers. For example, a distance can be a decimal, but it cannot be negative.


STEP: Decide if the answer can be negative
[−1 point ⇒ 0 / 1 points left]

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a negative number or not.

Focus on what the problem asks: what was the winner's average speed for the marathon. That means we need a speed value. Negative speed is not possible: it must be a positive number.

The correct answer is: No, it cannot be negative.


Submit your answer as:

Word Problems: checking answers

Suppose you must solve this word problem:

A marathon is a long-distance running race with an official distance of 42,195 kilometres. 1 000 people ran the "One Metre at a Time" marathon in Polokwane. The second-place runner finished 15,69 seconds after the winner. The third runner was 12,51 seconds behind the second runner, and the fourth runner was 29,41 seconds behind the third runner. If the second-place runner had an average speed of 18,57 km/h, how many of the runners took more than 2,36 hours to finish the marathon?

Can the answer to the question above be a negative number?

Answer:

Can the answer be negative?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Think about the meaning of the number you need. Some values cannot be decimal numbers. Other values cannot be negative numbers. For example, a distance can be a decimal, but it cannot be negative.


STEP: Decide if the answer can be negative
[−1 point ⇒ 0 / 1 points left]

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a negative number or not.

Focus on what the problem asks: how many of the runners took more than 2,36 hours to finish the marathon. So the question is about the number of people who finished the race after 2,36 hours had passed. The answer cannot be negative. The number of people to finish after 2,36 hours might be 3 or 7 people, but it cannot be 5 people. In other words, the number of people must be a positive number or zero, but it cannot be negative.

The correct answer is: No, it cannot be negative.


Submit your answer as:

Word Problems: checking answers

Suppose you must solve this word problem:

A marathon is a long-distance running race with an official distance of 42,195 kilometres. 1 000 people ran the "One Metre at a Time" marathon in Port Elizabeth. The second-place runner finished 15,46 seconds after the winner. The third runner was 14,66 seconds behind the second runner, and the fourth runner was 27,92 seconds behind the third runner. If the second-place runner had an average speed of 19,93 km/h, how many of these runners finished the marathon in less than 2,18 hours?

Can the answer to the question above be a negative number?

Answer:

Can the answer be negative?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Think about the meaning of the number you need. Some values cannot be decimal numbers. Other values cannot be negative numbers. For example, a distance can be a decimal, but it cannot be negative.


STEP: Decide if the answer can be negative
[−1 point ⇒ 0 / 1 points left]

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a negative number or not.

Focus on what the problem asks: how many of these runners finished the marathon in less than 2,18 hours. So the question is about the number of people who finished the race before 2,18 hours had passed. The answer cannot be negative. The number of people to finish before 2,18 hours might be 3 or 7 people, but it cannot be 5 people. In other words, the number of people must be a positive number or zero, but it cannot be negative.

The correct answer is: No, it cannot be negative.


Submit your answer as:

Word problems: rectangle facts

The diagonal of a rectangle is 32 cm more than its width. The length of the same rectangle is 16 cm more than its width.

What are the width and the length of the rectangle?

Answer: The width is cm and length is cm.
numeric
numeric
HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.


STEP: Draw a diagram and label it
[−1 point ⇒ 7 / 8 points left]

Drawing a diagram is a helpful way to organise the information in a question about a shape. This question is about a rectangle, so we can start with a rectangle picture. The question mentions the diagonal of the ractangle, so we should draw that also.

Notice we have added labels for the three parts of the rectangle mentioned in the question: d is the diagonal, w is the width, and l is the length.


STEP: Connect the variables in the diagram
[−2 points ⇒ 5 / 8 points left]

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

32 cm more than its widththe diagonal of the rectangle isd=w+3216 cm more than its widththe length of the rectangle isl=w+16

STEP: Use the Theorem of Pythagoras and solve the equations simultaneously
[−2 points ⇒ 3 / 8 points left]

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

From the picture we can see that there are two right-angled triangles in the rectangle. As always for a right-angled triangle, we can use the Theorem of Pythagoras. Using our variables from the diagram, we can write:

d2=w2+l2

And now we can substitute in the equations from above. Specifically we can substitute w+32 in for d and w+16 in for l. Then expand the binomials and simplify the equation as much as possible.

d2=w2+l2(w+32)2=w2+(w+16)2w2+64w+1 024=w2+(w2+32w+256)0=w232w768

STEP: Solve the equation for w
[−2 points ⇒ 1 / 8 points left]

We have a quadratic equation in standard form! It is time to solve it.

0=w232w7680=(w48)(w+16)w=48or w=16

This means that the width of the rectangle is either 48 cm or 16 cm. But the dimensions of a rectangle cannot be negative, so w=48 cm.


STEP: Calculate the length of the rectangle
[−1 point ⇒ 0 / 8 points left]

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects l to w.

l=w+16=(48)+16=64 cm

The width of the rectangle is 48 cm and the length is 64 cm.


Submit your answer as: and

Word problems: rectangle facts

The diagonal of a rectangle is 34 cm more than its width. The length of the same rectangle is 17 cm more than its width.

Determine the width and length of the rectangle.

Answer: The width is cm and length is cm.
numeric
numeric
HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.


STEP: Draw a diagram and label it
[−1 point ⇒ 7 / 8 points left]

Drawing a diagram is a helpful way to organise the information in a question about a shape. This question is about a rectangle, so we can start with a rectangle picture. The question mentions the diagonal of the ractangle, so we should draw that also.

Notice we have added labels for the three parts of the rectangle mentioned in the question: d is the diagonal, w is the width, and l is the length.


STEP: Connect the variables in the diagram
[−2 points ⇒ 5 / 8 points left]

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

34 cm more than its widththe diagonal of the rectangle isd=w+3417 cm more than its widththe length of the rectangle isl=w+17

STEP: Use the Theorem of Pythagoras and solve the equations simultaneously
[−2 points ⇒ 3 / 8 points left]

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

From the picture we can see that there are two right-angled triangles in the rectangle. As always for a right-angled triangle, we can use the Theorem of Pythagoras. Using our variables from the diagram, we can write:

d2=w2+l2

And now we can substitute in the equations from above. Specifically we can substitute w+34 in for d and w+17 in for l. Then expand the binomials and simplify the equation as much as possible.

d2=w2+l2(w+34)2=w2+(w+17)2w2+68w+1 156=w2+(w2+34w+289)0=w234w867

STEP: Solve the equation for w
[−2 points ⇒ 1 / 8 points left]

We have a quadratic equation in standard form! It is time to solve it.

0=w234w8670=(w51)(w+17)w=51or w=17

This means that the width of the rectangle is either 51 cm or 17 cm. But the dimensions of a rectangle cannot be negative, so w=51 cm.


STEP: Calculate the length of the rectangle
[−1 point ⇒ 0 / 8 points left]

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects l to w.

l=w+17=(51)+17=68 cm

The width of the rectangle is 51 cm and the length is 68 cm.


Submit your answer as: and

Word problems: rectangle facts

The diagonal of a rectangle is 32 cm more than its width. The length of the same rectangle is 16 cm more than its width.

What are the width and the length of the rectangle?

Answer: The width is cm and length is cm.
numeric
numeric
HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.


STEP: Draw a diagram and label it
[−1 point ⇒ 7 / 8 points left]

Drawing a diagram is a helpful way to organise the information in a question about a shape. This question is about a rectangle, so we can start with a rectangle picture. The question mentions the diagonal of the ractangle, so we should draw that also.

Notice we have added labels for the three parts of the rectangle mentioned in the question: d is the diagonal, w is the width, and l is the length.


STEP: Connect the variables in the diagram
[−2 points ⇒ 5 / 8 points left]

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

32 cm more than its widththe diagonal of the rectangle isd=w+3216 cm more than its widththe length of the rectangle isl=w+16

STEP: Use the Theorem of Pythagoras and solve the equations simultaneously
[−2 points ⇒ 3 / 8 points left]

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

From the picture we can see that there are two right-angled triangles in the rectangle. As always for a right-angled triangle, we can use the Theorem of Pythagoras. Using our variables from the diagram, we can write:

d2=w2+l2

And now we can substitute in the equations from above. Specifically we can substitute w+32 in for d and w+16 in for l. Then expand the binomials and simplify the equation as much as possible.

d2=w2+l2(w+32)2=w2+(w+16)2w2+64w+1 024=w2+(w2+32w+256)0=w232w768

STEP: Solve the equation for w
[−2 points ⇒ 1 / 8 points left]

We have a quadratic equation in standard form! It is time to solve it.

0=w232w7680=(w48)(w+16)w=48or w=16

This means that the width of the rectangle is either 48 cm or 16 cm. But the dimensions of a rectangle cannot be negative, so w=48 cm.


STEP: Calculate the length of the rectangle
[−1 point ⇒ 0 / 8 points left]

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects l to w.

l=w+16=(48)+16=64 cm

The width of the rectangle is 48 cm and the length is 64 cm.


Submit your answer as: and

Setting up simultaneous equations

Last week, Abodunrin and Babalwe had a chemistry test. Now they are comparing their marks and they notice these facts:

  • The sum of the marks is 152.
  • Abodunrin's mark is 22 more than Babalwe's mark.

Let a represent Abodunrin's mark and b represent Babalwe's mark. Then which equations below accurately represent the facts? Select your answer from the choices below.

Answer:
Fact about the test scores Equation
The sum of the marks is 152.
Abodunrin's mark is 22 more than Babalwe's mark.
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is more. Use these key words to change the words into operations, for example, addition or multiplication.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable a for Abodunrin's mark and b for Babalwe's mark. So we can break up the first fact like this:

The sum of the marksis152a+b=152

The first fact is equivalent to this equation: a+b=152.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

Similarly, we can identify key parts of the second fact, and translate each into an expression.

Abodunrin's markis22 more than Babalwe's marka=b+22

This equation, a=b+22, means that Abodunrin's mark is more that Babalwe's mark.

The correct answers are:

Fact about the test scores Equation
The sum of the marks is 152. a+b=152
Abodunrin's mark is 22 more than Babalwe's mark. a=b+22

Submit your answer as: and

Setting up simultaneous equations

Last week, Abdoul and Balarabe had a maths test. Now they are comparing their marks and they notice these facts:

  • The sum of the marks is 126.
  • Abdoul's mark is 6 more than Balarabe's mark.

Let a represent Abdoul's mark and b represent Balarabe's mark. Then which equations below accurately represent the facts? Select your answer from the choices below.

Answer:
Fact about the test scores Equation
The sum of the marks is 126.
Abdoul's mark is 6 more than Balarabe's mark.
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is more. Use these key words to change the words into operations, for example, addition or multiplication.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable a for Abdoul's mark and b for Balarabe's mark. So we can break up the first fact like this:

The sum of the marksis126a+b=126

The first fact is equivalent to this equation: a+b=126.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

Similarly, we can identify key parts of the second fact, and translate each into an expression.

Abdoul's markis6 more than Balarabe's marka=b+6

This equation, a=b+6, means that Abdoul's mark is more that Balarabe's mark.

The correct answers are:

Fact about the test scores Equation
The sum of the marks is 126. a+b=126
Abdoul's mark is 6 more than Balarabe's mark. a=b+6

Submit your answer as: and

Setting up simultaneous equations

Last week, Adanna and Bokamoso had a physics test. Now they are comparing their marks and they notice these facts:

  • The sum of the marks is 159.
  • Adanna's mark is 7 more than Bokamoso's mark.

Let a represent Adanna's mark and b represent Bokamoso's mark. Then which equations below accurately represent the facts? Select your answer from the choices below.

Answer:
Fact about the test scores Equation
The sum of the marks is 159.
Adanna's mark is 7 more than Bokamoso's mark.
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is more. Use these key words to change the words into operations, for example, addition or multiplication.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable a for Adanna's mark and b for Bokamoso's mark. So we can break up the first fact like this:

The sum of the marksis159a+b=159

The first fact is equivalent to this equation: a+b=159.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

Similarly, we can identify key parts of the second fact, and translate each into an expression.

Adanna's markis7 more than Bokamoso's marka=b+7

This equation, a=b+7, means that Adanna's mark is more that Bokamoso's mark.

The correct answers are:

Fact about the test scores Equation
The sum of the marks is 159. a+b=159
Adanna's mark is 7 more than Bokamoso's mark. a=b+7

Submit your answer as: and